Solved Problems

(a) First, find each of the electric fields, due to the charges, at P and then apply vector addition to calculate the resultant electric field at that point. We know that the electric field due to a point charge at distance $r$ from the source is given by
\[\vec E=\frac{1}{4\pi \epsilon_0} \frac{\left|q\right|}{r} \hat r\]
Where $\hat r$ is the unit vector lie along the line joining the source and the desired location of the field point. In some literature $\frac{1}{4\pi \epsilon_0}$ is called $k$. Therefore, \begin{gather*} {\vec{E}}_1=\frac{k\left|q_1\right|}{r^2_1}{\hat{r}}_1=\frac{kQ}{a^2}\left({\cos 60{}^\circ}\hat{i}+{\sin 60{}^\circ}(-\hat{j})\right)\\\\ {\vec{E}}_2=\frac{k\left|q_2\right|}{r^2_2}{\hat{r}}_2=\frac{k\left(2Q\right)}{a^2}(-\hat{i})\\\\ {\vec{E}}_3=\frac{k\left|q_3\right|}{r^2_3}{\hat{r}}_3=\frac{kQ}{a^2}({\cos 60{}^\circ}\hat{i}+{\sin 60{}^\circ}\hat{j})\\\\ {\vec{E}}_P={\vec{E}}_1+{\vec{E}}_2+{\vec{E}}_3=\frac{kQ}{a^2}2({\cos 60{}^\circ}-1)\hat{i}=-\frac{kQ}{a^2}\hat{i}\end{gather*}

(b) The force exerted on a charge $q$ in a uniform electric field is given by $\vec{F}=q\vec{E}$. 
\[\vec{F}=q\vec{E}=\left(-e\right)\left(-\frac{kQ}{a^2}\right)\hat{i}=\frac{keQ}{a^2}\hat{i}\] 

Note: the electric field of a negative point charge is toward the charge and away from the positive charge.

Recall that the electric potential due to a system of point charges at an arbitrary point is defined as $V={\Sigma }^N_{i=1}V_i={\Sigma }^N_{i=1}k\frac{q_i}{r_i}$ . Therefore first find the distance of each point charges to the point P and then find the electric potential. 
\[r_3=\sqrt{r^2_{23}-r^2_2}=1.16\ {\rm m}\] 
\[r_2=\frac{1}{2}r_{12}=0.67\ {\rm m\ \ ,\ \ }{{\rm r}}_{{\rm 1}}{\rm =}\frac{{\rm 1}}{{\rm 2}}r_{12}=0.67\ {\rm m}\] 

\[V_P=k\left(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}+\frac{Q_3}{r_3}\right)\] 
\[=\left(8.99\times {10}^9\right)\left(\frac{6.84\times {10}^{-6}}{0.67}+\frac{\left(-16.2\times {10}^{-6}\right)}{0.67}+\frac{9.20\times {10}^{-6}}{1.16}\right)\] 
\[=\left(8.99\times {10}^9\right)\left(-6.04\times {10}^{-6}\right)=-5.43\times {10}^4\ {\rm V}\] 
 

Two forces acting on the second proton. One is the Coulomb force due to the first proton and the second is the gravity so these forces must be in balance. 
\begin{align*} F_e&=F_g \\ \\\frac{kq_1q_2}{r^2}&=m_2g\\\\ \Rightarrow r&=\sqrt{\frac{kq_1q_2}{m_2g}}\\\\&=\sqrt{\frac{\left(9\times {10}^9\right){\left(1.06\times {10}^{-19}\right)}^2}{\left(1.672\times {10}^{-27}\right)\left(9.8\right)}}\\\\&=0.119\quad{\rm m}\end{align*}

Note: the force acting on a positive charge due to an electric field is in direction of the field but a negative charge is in opposite direction. So $E$ must be pointed to the left (see the figure).

Since the system is in equilibrium so apply the equilibrium conditions to one of the spheres, say the negative sphere, i.e. $\Sigma F_y=0\ ,\ \Sigma F_x=0$
\begin{gather*} \Sigma F_y=T{\cos  10{}^\circ \ }-mg=0\\\\ \Rightarrow T=\frac{mg}{{\cos  10{}^\circ \ }}=\frac{\left(0.002\right)\left(9.8\right)}{{\cos  10{}^\circ \ }}=0.02\, {\rm N} \\ \\ \Sigma F_x=\underbrace{T{\sin  10{}^\circ \ }}_{T_x}+\underbrace{\frac{kq^2}{r^2}}_{F_e}-qE=0\end{gather*} Therefore, \begin{align*}E&=\frac{1}{5\times {10}^{-8}}\left((0.02)(\sin 10^\circ)+\frac{(9\times {10}^9){\left(5\times {10}^{-8}\right)}^2}{(0.03473)^2}\right)\\\\&=4.42\times {10}^5\quad {\rm N/C}\end{align*} Where $r$ is the distance between the charges and is found by Pythagoras theorem as shown in the figure. \begin{gather*} \sin 10^\circ=\frac{\frac{r}{2}}{10}\\\\\Rightarrow r=20\,{\sin 10^\circ}=3.473\quad {\rm cm}\end{gather*}

Superposition principle: consider a sphere with charge density $\rho$ and a sphere with charge density $-\rho$ at point $\vec{a}$ as shown in the figure below. The radius of the sphere with $\rho$ is larger than that of $-\rho$. Now using Gauss's law, we must find the electric field inside a uniform charge density at arbitrary point as
\[\oint_S{\vec{E}.d\vec{S}}=\frac{Q_{enc}}{\epsilon_0}\Rightarrow \vec{E}.\oint_S{d\vec{S}}=\frac{\rho V_{Gaussian}}{\epsilon_0}\] 
Where $\oint_S{d\vec{S}}$ is the area of the Gaussian surface with radius $r$ inside of the sphere with radius $R$. So the electric field at distance $r$ from the center of sphere is 
\[E\left(4\pi r^2\right)=\frac{\rho\left(\frac{4}{3}\pi r^3\right)}{\epsilon_0}\Rightarrow \vec{E}=\frac{\rho}{3\epsilon_0}\vec{r}\] 
In this problem we have two spheres with charge densities $+\rho$ and $-\rho$ . We want to find the electric field due to these charges at point P.

\[{\vec{E}}_{+\rho}(P)=+\frac{\rho}{3\epsilon_0}\vec{r}\] 
\[{\vec{E}}_{-\rho}(P)=\frac{-\rho}{3\epsilon_0}(\vec{r}-\vec{a})\] 
\[{\vec{E}}_P={\vec{E}}_{+\rho}\left(P\right)+{\vec{E}}_{-\rho}\left(P\right)=+\frac{\rho}{3\epsilon_0}\vec{r}-\frac{\rho}{3\epsilon_0}\left(\vec{r}-\vec{a}\right)\] 
\[\Rightarrow {\vec{E}}_P=+\frac{\rho}{3\epsilon_0}\vec{a}\] 
 

By definition, $E=-dV(x)/dx\ $, therefore 
\[E=-\frac{d}{dx}\left(\frac{6}{x^2}\right)=-\frac{d}{dx}\left(-12x^{-3}\right)=\frac{12}{x^3}\] 
 

The potential electric at the distance $r$ from the point charge $q$ is determined by $V\left(r\right)=kq/r$. The potential difference of a collection of point charges at arbitrary point A is found by summing individual potential at that point as $V_A=V_1+V_2+\dots $. Therefore, 
\[V_A=V_4+V_{-4}=\frac{kq_1}{r_1}+\frac{kq_2}{r_2}=\left(9\times {10}^9\right)\left\{\frac{4\times {10}^{-6}}{0.3}+\frac{-4\times {10}^{-6}}{\sqrt{{\left(0.4\right)}^2+{\left(0.3\right)}^2}}\right\}=48\ {\rm kV}\] 
\[V_B=V_4+V_{-4}=\frac{kq_1}{r_4}+\frac{kq_2}{r_3}=\left(9\times {10}^9\right)\left\{\frac{4\times {10}^{-6}}{0.5}+\frac{-4\times {10}^{-6}}{0.3}\right\}=-48\ {\rm kV}\] 
So, $V_A-V_B=48-\left(-48\right)=96\ {\rm kV}$

By definition, the potential difference between two points A and B are 
\[\Delta V=V_B-V_A=-\int^B_A{\vec{E}.d\vec{l}}\] 
Since the electric field in this case is uniform so can be factored from the integral as

$\Delta V=-\vec{E}.\int^B_A{d\vec{l}}=-\vec{E}.\Delta \vec{r}$ , where $\Delta \vec{l}$ is the displacement vector between A, B and defined as $\Delta \vec{r}={\vec{r}}_B-{\vec{r}}_A=\left(x_B-x_A\right)\hat{i}+\left(y_B-y_A\right)\hat{j}=5\hat{i}+\left(-5\right)\hat{j}$. Therefore,
\[\Delta V=-\left(500\hat{i}-200\hat{j}\right).\left(5\hat{i}-5\hat{j}\right)=-\left[\left(500\times 5\right)\underbrace{\left(\hat{i}.\hat{i}\right)}_{1}+\left(200\times 5\right)\underbrace{\left(\hat{j}.\hat{j}\right)}_{1}\right]\] 
\[\therefore \Delta V=-3500\ {\rm V}\] 
 

As previous, $\Delta V=V_B-V_A=-\int^B_A{\vec{E}.d\vec{l}}\ $where $d\vec{l}$ is the displacement vector between the points so that in this case $d\vec{l}=dr\ \hat{r}$. By Gauss's theorem, the electric field of an infinitely long thin wire is determine as $\vec{E}=\frac{2k\lambda}{r}\hat{r}$, therefore 
\[\Delta V=V_5-V_6=-\int^5_6{\frac{2k\lambda}{r}dr}=-2k\lambda{\left.{\ln  r\ }\right|}^5_6=-2k\lambda{\ln  \frac{5}{6}\ }\] 
\[V_5-V_6=-2\left(9\times {10}^9\right)\left(264\times {10}^{-9}\right){\ln  \frac{5}{6}\ }=-0.866\ {\rm V}\] 

\[\therefore \ \left|\Delta V\right|=0.866\ {\rm V}\] 
 

First, find the electric force due to each charge on the $q_3$, then use the superposition principle to do their vector sum.

\begin{align*} {\vec{F}}_{13}&=k\frac{\left|q_1\right|\left|q_3\right|}{r^2_{13}}{\hat{r}}_{13}\\&=\left(9\times {10}^9\right)\frac{\left(0.71\times {10}^{-6}\right)\left(1\times {10}^{-6}\right)}{(0.10)^2}\left({\cos \theta}\hat{x}+{\sin \theta}(-\hat{y})\right)\end{align*}
Where we have decomposed the unit vector as above.

Since $q_1>0$ so the electric field lines are along the line between $q_1$ and $q_3$ and directed away from $q_3$. From the geometry we see that ${\sin  \theta\ }=\frac{8}{10}$ and ${\cos  \theta\ }=\frac{\sqrt{{10}^2-8^2}}{10}=\frac{6}{10}$. Therefore,
\begin{align*} {\vec{F}}_{13}&=0.639(0.8\,\hat{x}+0.6\,(-\hat{y}))\\&=(0.511\,\hat{x}-0.383\,\hat{y}){\rm \quad N}\end{align*} Now find the electric force due to the $q_2$ on $q_3$ i.e. ${\vec{F}}_{23}$ \begin{align*}{\vec{F}}_{23}&=k\frac{\left|q_2\right|\left|q_3\right|}{r^2_{23}}{\hat{r}}_{23}\\&=\left(9\times {10}^9\right)\frac{\left|-0.67\times {10}^{-6}\right|\left(1\times {10}^{-6}\right)}{\left({10}^2-8^2\right)\times {10}^{-4}}(-\hat{y})\\&=(-1.675\, \hat{y})\quad {\rm N}\end{align*} Therefore, the resultant force on the $q_3$ is \begin{align*}{\vec{F}}_3&={\vec{F}}_{13}+{\vec{F}}_{23}\\\\&=0.511\,\hat{x}-0.383\,\hat{y}+(-1.675\, \hat{y})\\\\&=(0.511\, \hat{x}-2.058\, \hat{y})\quad {\rm N}\end{align*} The direction of the net force with the $x$ axis are determined by ${\tan \alpha}=\left|F_y/F_x\right|$, so \[\alpha=\tan^{-1} \left(\frac{2.058}{0.511}\right)=76.05{}^\circ \] 
Since $F_{3x}>0$ and $F_{3y}<0\ $so the net force lies in the fourth quadrant.
And its magnitude is \[|{\vec{F}}_3|=\sqrt{(0.511)^2+(-2.058)^2}=2.12\quad {\rm N}\]

There is three forces acting on the $q_2$. The electric force due to $q_1$, tension in the thread and the gravity. Thus, its free body diagram is as follows

The system is in equilibrium so the net force on the $q_2$ is zero i.e. ${\left(\Sigma F_y\right)}_2=0$
\[\Rightarrow F_e-T-mg=0\ \Rightarrow T=\frac{k\left|q_1\right|\left|q_2\right|}{{\left(15\right)}^2}-mg\] 
\[\Rightarrow T=9\times {10}^9\frac{\left(2\times {10}^{-6}\right)\left(2\times {10}^{-6}\right)}{{\left(0.15\right)}^2}-\left(0.040\times 9.8\right)=1.208\ {\rm N}\] 
 

Because the net electric field at origin is zero, so we have
\[{\vec{E}}_O={\vec{E}}_Q+{\vec{E}}_1+{\vec{E}}_2=0\] 
\[\frac{k\left|-Q\right|}{r^2}\hat{r}+\frac{k\left|q_1\right|}{r^2_1}{\hat{r}}_1+\frac{k\left|q_2\right|}{r^2_2}{\hat{r}}_2=0\] 
\[\frac{kQ}{2^2}\left({\cos  30{}^\circ \ }\hat{x}+{\sin  30{}^\circ \ }\hat{y}\right)+\frac{k\left|q_1\right|}{{\left(1.1\right)}^2}{\hat{r}}_1+\frac{k\left|q_2\right|}{{\left(1.3\right)}^2}{\hat{r}}_2=0\] 

Since $q_1$ and $q_2$ are located in the $x$ and $y$ axis, respectively, so the directions of the electric fields are along theses axes and in opposite direction of the $x$ and $y$ components of the $\vec E_Q$ since the net electric field at origin must be zero. . i.e. ${\vec{E}}_1=\left|E_1\right|\hat{-x}$ and ${\vec{E}}_2=\left|E_2\right|\hat{-y}$. Therefore,
\[\frac{\left(500\times {10}^{-9}\right)}{4}{\cos  30{}^\circ \ }-\frac{\left|q_1\right|}{1.21}=0\Rightarrow \ \left|q_1\right|=1.31\times {10}^{-7}\ {\rm C=131\ nC}\] 
\[\frac{\left(500\times {10}^{-9}\right)}{4}{\sin  30{}^\circ \ }-\frac{\left|q_2\right|}{{\left(1.3\right)}^2}=0\Rightarrow \left|q_2\right|=1.05\times {10}^{-7}{\rm C}=105\ {\rm nC}\]

Thus the charge $q_1$ is positive and charge $q_2$ is negative (since the direction of the electric field is toward the charge $q_2$ and away from charge $q_1$ )

This configuration is a parallel plate capacitor. Recall that in this case the electric field between the plates is determined by $E=\sigma {\rm /}\epsilon_0$, where $\sigma$ is the surface charge density of the plates. Therefore,
\[\sigma=\epsilon_0E=\frac{Q}{A_{disk}}\Rightarrow \ \epsilon_0E=\frac{Q}{\pi r^2}\Rightarrow r=\sqrt{\frac{Q}{\pi \epsilon_0E}}\] 
\[\Rightarrow r=\sqrt{\frac{ne}{\pi \epsilon_0E}}=\sqrt{\frac{\left(3.9\times {10}^9\right)\left(1.6\times {10}^{-19}\right)}{\pi\left(8.85\times {10}^{-12}\right)\left({10}^5\right)}}=0.015\ {\rm m}\] 
In above we have used the relation $Q=ne$, where $n$ is the number of the electrons.
 

Note: the electric field due to an infinitely long line charge of uniform charge density $\lambda$ at distance $r$ from it is ${\vec{E}}_r=\frac{1}{2\pi\epsilon_0}\frac{\lambda}{r}\hat{r}=2k\frac{\lambda}{r}\hat{r}$. Therefore, 
\[\Rightarrow \lambda=\frac{rE}{2k}=\frac{0.058\times 2000}{2\times 9\times {10}^9}=6.44\times {10}^{-9}\frac{{\rm C}}{{\rm m}}\] 
By definition, linear charge density is 
\[\lambda=\frac{Q}{L}\Rightarrow Q=\lambda L=\left(6.44\times {10}^{-9}\right)\left(0.01\right)=-0.0644\ {\rm nC}\] 
Since the fields line are toward the wire so the distributed charge is negative!
 

(a) Recall that the electric field of a point charge at distance $r$ is $\vec{E}=k\frac{\left|q\right|}{r^2}\hat{r}$. So first find the electric fields of the charges at P

\begin{align*} {\vec{E}}_2&=k\frac{\left|q_2\right|}{r^2_2}{\hat{r}}_2\\&=k\frac{\left|-2\times {10}^{-9}\right|}{{\left(0.03\right)}^2}\left(-\hat{y}\right)\\ &=22.22\times {10}^{-9}k\ (-\hat{y}) \end{align*} Since ${\vec{E}}_2$ is vertical and points in the negative $y$ direction (for negative point charges, the field lines are directed radially inward), we have chosen $(-\hat{y})$.
\begin{align*}{\vec{E}}_1&=k\frac{|q_1|}{r^2_1}{\hat{r}}_1\\\\&=k\frac{2.5\times {10}^{-9}}{(0.04)^2+(0.03)^2}{\hat{r}}_1\\\\&=(10\times {10}^{-9})k({\cos \theta}\,\hat{x}+{\sin \theta}\,\hat{y}) \end{align*} Where we have decomposed the unit vector in the $r_1$ direction. (Since $q_1$ is positive the electric field lines points directly away from the point P and vice versa) 
From the geometry \[{\sin  \theta\ }=\frac{3}{5}\ \ ,\ {\cos  \theta\ }=\frac{4}{5}\ \] To find the total electric field at P combine the results above as\begin{align*}{\vec{E}}_P&={\vec{E}}_1+{\vec{E}}_2\\\\&=(10\times {10}^{-9})k\left(\frac{4}{5}\hat{x}+\frac{3}{5}\ \hat{y}\right)+22.22\times {10}^{-9}k\,(-\hat{y})\\\\&={10}^{-9}k(8\hat{x}-16.22\hat{y})\end{align*}The magnitude of ${\vec{E}}_P$ is as follows \begin{align*}\left|{\vec{E}}_P\right|&=\sqrt{E^2_x+E^2_y}\\\\&=\sqrt{{\left(8\times {10}^{-9}k\right)}^2+{\left(-16.22\times {10}^{-9}k\right)}^2}\\\\&=18.08\times {10}^{-9}k\\\\&=162.58\quad {\rm N/C}\end{align*} Where we have substituted $k=8.99\times {10}^9{\rm \frac{N.m^2}{C^2}}$

(b) The electric potential due to a point charge $q$ at distance $r$ is defined as $V(r)=k\frac{q}{r}$ \begin{gather*} V_1\left(P\right)=k\frac{q_1}{r_1}=k\frac{-2\times {10}^{-9}}{0.03}=-599.3\quad {\rm V}\\\\ V_2(P)=k\frac{q_2}{r_2}=\left(8.99\times {10}^9\right)\frac{2.5\times {10}^{-9}}{0.05}=449.5\quad {\rm V} \\\\ \therefore \quad V_P=V_1+V_2=-149.8\ {\rm V}\end{gather*}
(c) If a charge is located in a uniform electric field $E$, then a force with a magnitude of $F=qE$ exerts on it.
\begin{align*}F_P&=qE_P\\\\&=\left(2\times {10}^{-9}\right)\left(162.58\right)\\\\&=325.16\quad {\rm nN}\end{align*}

see the work-energy theorem as $\Delta K=K_f-K_i=W_{net}$, where $W_{net}$ is the net work done on the electron. The electric force acting on the electron in a uniform electric field is $F_e=eE$ and subsequently the work done on it is $W=Fx{\cos  \theta\ }=\left(eE\right)x{\cos  180{}^\circ \ }=-eEx$. (Since electron moves in the opposite direction of the field so we have $\theta=180{}^\circ $).
\[K_f-K_i=W_e\Rightarrow \frac{1}{2}mv^2_f-0=-\underbrace{\left(-e\right)}_{{\rm have\ negative\ charge}}Ex=eEx\] 
But recall that $\Delta V=Ex$. Therefore, 
\[\frac{1}{2}mv^2_f=e\Delta V\Rightarrow v_f=\sqrt{\frac{2e\Delta V}{m}}=\sqrt{\frac{2\left(1.6\times {10}^{-19}{\rm C}\right)\left(1300\right)}{9.10\times {10}^{-31}}}=2.138\times {10}^7\ {\rm m/s}\] 
 

First solution: use the work-energy theorem 

SI unit of energy is Joule so we must first convert initial kinetic energy into Joule
\[56.6\ {\rm MeV}\to {\rm 56.4\times }{{\rm 10}}^{{\rm 6}}\times \left(1.6\times {10}^{-19}\right)=9.035\times {10}^{-12}{\rm J}\] 
\[\Delta K=K_2-K_1=W_{net}\] 
\begin{align*}
0-K_1=\left(qE\right)x\,{\cos  180{}^\circ \ }\Rightarrow E &=\frac{K_1}{qx}=\frac{9.035\times {10}^{-12}}{2\left(1.6\times {10}^{-19}\right)\left(6620\times {10}^3\right)}\\
&= 4.26\ \frac{{\rm N}}{{\rm C}}
\end{align*}
Second solution: first using $v^2-v^2_0=2a\Delta x$ find the acceleration of the particle, then by the Newton's 2${}^{nd}$ law determine the force exerted on it.
\[K_0=\frac{1}{2}\underbrace{m}_{4m_p}v^2_0\Rightarrow v_0=\sqrt{\frac{2K_0}{4m_p}}=\sqrt{\frac{2\left(9.035\times {10}^{-12}\right)}{6.688\times {10}^{-27}}}=5.198\times {10}^7\frac{{\rm m}}{{\rm s}}\] 
\[v^2-v^2_0=2a\Delta x\Rightarrow 0-{\left(5.1987\times {10}^7\right)}^2=2a\left(6620\times {10}^3\right)\Rightarrow a=-2.04\times {10}^8\frac{{\rm m}}{{{\rm s}}^{{\rm 2}}}\] 
 The minus sign indicates that the object is going to decrease its velocity.
\[F=ma=4m_pa=6.688\times {10}^{-27}\times \left(-2.04\times {10}^8\right)=-1.36\times {10}^{-18}\ {\rm N}\] 
By definition, the magnitude of the electric field is 
\[E=\frac{\left|F\right|}{q}=\frac{\left|-1.36\times {10}^{-18}\right|}{2(1.6\times {10}^{-19})}=4.26\frac{{\rm N}}{{\rm C}}\] 
 

When a particle moves in an electric field, experiences a force by the field as $F=qE$. Using the work-energy theorem, we get
\[\Delta K=K_2-K_1=W_{net}\] 
\[0-\frac{1}{2}mv^2_0=\left(qE\right)x\,{\cos  180{}^\circ \ }\Rightarrow \frac{q}{m}=\frac{v^2_0}{2Ex}\] 
\[\Rightarrow \frac{q}{m}=\frac{{\left(4\times {10}^5\right)}^2}{2\times 100\times 1.2\times {10}^{-2}}=6.67\times {10}^{10}\frac{{\rm C}}{{\rm kg}}\] 
Note: since the particle comes to a stop so there is a dissipative force exerted on it that is in opposite direction to the motion of the particle so $W=Fx{\cos  180{}^\circ \ }$.

When a positive charge enters in a region with constant $E$ then it moves in the direction of the field and its velocity increased but a negative charge moves opposite of the field and therefore its velocity decreased until come to a stop. In this problem the charge after entering the field come to a stop so it must be negative charge!
 

Given data: $v_0=0\ ,\ E=5000\frac{{\rm N}}{{\rm C}}\ ,\ \ \ v=100\frac{{\rm m}}{{\rm s}}\ ,\ t=?$
From the kinematic relation $v=v_0+at$ we need to find the acceleration of the electron in the electric field to calculate the desired time. By definition, force is related to the electric field via $F=qE$ hence
\[F=qE\Rightarrow ma=qE\Rightarrow a=\frac{qE}{m}\] 
Substituting in the above kinematic relation and solving for $t$, gets
\[v=v_0+at=v_0+\left(\frac{q}{m}\right)Et\Rightarrow 100=0+\left(\frac{1.6\times {10}^{-19}}{9.11\times {10}^{-31}}\right)\left(5000\right)t\] 
\[\Rightarrow t=1.01\times {10}^{-13}\ {\rm s}\] 
 

(a) By definition, the linear charge density of a continuous line is $\lambda=dq/dx$. So
\begin{align*} 
dq=\lambda dx\to q_{tot}=\int{\lambda dx}&=2\int^{\frac{\pi}{2}}_0{\lambda_0\,{\cos \theta\ }\left(R\,d\theta\right)}\\
&=2\lambda_0R\,{\left({\sin \theta\ }\right)}^{\frac{\pi}{2}}_0\\
&=2\lambda_0R
\end{align*}
In the above, we have used the symmetry of the problem. 
(b) First, determine the electric field due to a charge element at P then integral over it to find the total electric field.
\begin{align*}
{\vec{E}}_P=\int{k\frac{dq}{R^2}\hat{r}}&=\int{k\frac{dq}{R^2}\left({\sin \theta\ }\left(-\hat{x}\right)+{\cos  \theta\ }\left(-\hat{y}\right)\right)}\\
&=\underbrace{\int{k\frac{dq}{R^2}{\sin  \theta\ }(-\hat{x})}}_{d{\vec{E}}_{Px}}+\underbrace{\int{k\frac{dq}{R^2}{\cos \theta\ }(-\hat{y})}}_{d{\vec{E}}_{Py}}\
\end{align*}
By symmetry, the $x$ component of the electric field becomes zero i.e. $d{\vec{E}}_{Px}=0$. So 
\begin{align*}
{\vec{E}}_P&=2\int^{\frac{\pi}{2}}_0{k\frac{dq}{R^2}\,{\cos \theta\ }(-\hat{y})}\\
&=2\frac{k}{R^2}\int^{\frac{\pi}{2}}_0{\lambda ds{\cos \theta\ }\left(-\hat{y}\right)}\\
&=2\frac{k}{R^2}\int^{\frac{\pi}{2}}_0{\left(\lambda_0{\cos  \theta\ }\right)(Rd\theta)\,{\cos \theta\ }\left(-\hat{y}\right)}\\
&= \frac{2k\lambda_0}{R}\int^{\frac{\pi}{2}}_0{{{\cos }^{{\rm 2}} \theta\ }d\theta\left(-\hat{y}\right)}\\
& =\frac{2k\lambda_0}{R}\int^{\frac{\pi}{2}}_0{\frac{1+{\cos  2\theta\ }}{2}}=\frac{2k\lambda_0}{R}\frac{1}{2}{\left(\theta+\frac{1}{2}{\sin  2\theta\ }\right)}^{\frac{\pi}{2}}_0\ \\
& =\frac{\pi k \lambda_0}{2R}\left(-\hat{y}\right)
\end{align*}

From the geometry below, we have 
First, calculate the electric field due to an element of charge $dq$ at the point $O$
\[d{\vec{E}}_O=k\frac{dq}{R^2}\hat{r}\xrightarrow{decompose\ \hat r} k\frac{dq}{r^2}\left({\sin \theta\ }\left(-\hat{x}\right)+{\cos  \theta\ }\left(\hat{y}\right)\right)\] 
By symmetry, we conclude that the $y$ component of the electric field is zero i.e. $d{\vec{E}}_{oy}=0$
So $d{\vec{E}}_O=dE_{ox}=-k\frac{dq}{R^2}{\sin  \theta\ }\hat{x}$.
Second, by integration of both sides of the above equation we can get the total electric field at the point P:
\begin{align*}
{\vec{E}}_O=-k\int{\frac{dq}{R^2}{\sin  \theta\ }}\hat{x} &=-k\int{\frac{\lambda Rd\theta}{R^2}\,{\sin \theta\ }}\hat{x}\\
&=-\frac{k\lambda}{R}\int^\pi_0{{\sin \theta\ }d\theta}\hat{x}\\
&=-\frac{k\lambda}{R}{\left(-{\cos  \theta\ }\right)}^\pi_0\\
&=-\frac{2k\lambda}{R}\hat{x}
\end{align*}
In above we have used the definition of the continuous charge density on a line i.e. 
$\lambda=\frac{dq}{dx}$ and the line element in a polar coordinate i.e. $dx=ds=Rd\theta$.
Now we can convert $\lambda$ into the explicit charge via $\lambda=\frac{q_{tot}}{L_{semicircle}}=\frac{q_{tot}}{\pi R}$
\begin{align*}
{\vec{E}}_O=-\frac{2k\lambda}{R}\hat{x}&=-\frac{2k\left(\frac{q}{L}\right)}{\frac{L}{\pi}}\hat{x}\\
&=-2k\frac{q\pi}{L^2}\hat{x}\\
&=-2\left(9\times {10}^9\right)\frac{\left(7.5\times {10}^{-6}\right)\pi}{{\left(0.14\right)}^2}\hat{x}\\
&=2.16\times {10}^7\left(-\hat{x}\right)\ \frac{{\rm N}}{{\rm C}}
\end{align*}
Therefore the total electric field at the point P is in the ($-x$) direction.
Finally, by definition of electric field $\vec{E}=\frac{\vec{F}}{q}$ we have 
\[F_o=qE_o=\left(3\times {10}^{-6}\right)\left(2.16\times {10}^7\right)\left(-\hat{x}\right)=64.8\ \left(-\hat{x}\right)\ {\rm N}\] 
 

First, calculate the electric fields due to the point charges at P then apply superposition principle: ${\vec{E}}_P={\vec{E}}_1+{\vec{E}}_2+{\vec{E}}_3$ to find the net electric field at that point.
\[\left|E_1\right|=\left|E_2\right|=k\frac{q}{r^2}=k\frac{q}{a^2+{\left(2a\right)}^2}=k\frac{q}{5a^2}\] 
\[{\vec{E}}_1=\left|E_1\right|{\hat{r}}_1=k\frac{q}{5a^2}\ {\hat{r}}_1=k\frac{q}{5a^2}\,\left({\cos \theta\ }\left(-\hat{x}\right)+{\sin \theta\ }\left(\hat{y}\right)\right)\] 
\[{\vec{E}}_2=\left|E_2\right|{\hat{r}}_2=k\frac{q}{5a^2}{\hat{r}}_2=k\frac{q}{5a^2}\,\left({\cos \theta\ }(-\hat{x})+{\sin \theta\ }\left(-\hat{y}\right)\right)\] 
But from the geometry 
\[{\cos  \theta\ }=\frac{2a}{r}=\frac{2a}{\sqrt{a^2+{\left(2a\right)}^2}}=\frac{2}{\sqrt{5}}\] 
\[{\sin  \theta\ }=\frac{a}{r}=\frac{a}{\sqrt{a^2+{\left(2a\right)}^2}}=\frac{1}{\sqrt{5}}\] 
Therefore 
\[{\vec{E}}_1=k\frac{q}{5a^2}\left(-\frac{2}{\sqrt{5}}\hat{x}+\frac{1}{\sqrt{5}}\hat{y}\right)\ \ ,\ \ {\vec{E}}_2=k\frac{q}{5a^2}\left(-\frac{2}{\sqrt{5}}\hat{x}-\frac{1}{\sqrt{5}}\hat{y}\right)\] 
On the other hand ${\vec{E}}_3=k\frac{2q}{{\left(2a\right)}^2}=k\frac{q}{2a^2}\hat{x}$
\[\therefore \ {\vec{E}}_P=k\frac{q}{5a^2}\left(-\frac{2}{\sqrt{5}}\hat{x}+\frac{1}{\sqrt{5}}\hat{y}\right)+k\frac{q}{5a^2}\left(-\frac{2}{\sqrt{5}}\hat{x}-\frac{1}{\sqrt{5}}\hat{y}\right)+k\frac{q}{2a^2}\hat{x}=k\frac{q}{a^2}\left(-\frac{4}{\sqrt{5}}+\frac{1}{2}\right) \hat{x}\ {\rm N/C\ }\]

The free body diagram below shows the external forces acts on the ball:
Because the ball is in equilibrium (the ball is motionless) so first write out the equilibrium conditions in $x$ and $y$ directions as follows
\[\Sigma F_x=0=\Sigma F_y\] 
\[\Sigma F_y=0\Rightarrow T{\cos  \theta\ }-F_g=mg\] 
\[\Sigma F_x=0\Rightarrow F_e=T{\sin \theta\ }\] 
Dividing both relations by each other, we obtain the angle of the equilibrium position: 
\[\therefore {\tan  \theta\ }=\frac{{\sin  \theta\ }}{{\cos \theta\ }}=\frac{\frac{F_e}{T}}{\frac{mg}{T}}=\frac{F_e}{mg}=\frac{qE}{mg}\] 
Solving for $q$, we get
\[q=\frac{mg}{E}{\tan  \theta\ }=\frac{0.002\times 9.8}{{10}^3}{\tan  15{}^\circ \ }=5.25\ \mu{\rm C}\] 

Because the magnitude and distance of all of the charges are equal so 
\[\left|F_1\right|=\left|F_2\right|=\left|F_3\right|=\left|F_4\right|=k\frac{\left|qQ\right|}{r^2}\] 
By symmetry consideration, $F_x=F_y=0$. So the direction of one of the forces is:
\[{\vec{F}}_{1z}=k\frac{\left|qQ\right|}{r^2}\,{\cos \theta\ }\left(-\hat{k}\right)=-k\frac{\left|qQ\right|}{r^3}z\hat{k}\] 
Where we have used from the geometry of the problem ${\cos \theta\ }=z/r$.  

By symmetry ${\vec{F}}_{1z}={\vec{F}}_{2z}={\vec{F}}_{3z}={\vec{F}}_{4z}=-k\frac{\left|qQ\right|}{r^3}z\hat{k}$

So
\[{\vec{F}}_z={\Sigma }^4_{i=1}{\vec{F}}_{iz}=-4k\frac{qQ}{r^3}z\hat{k}\] 
In terms of the parameters of the square and using the Pythagoras theorem, we have:
\[x=\frac{\sqrt{2}}{2}L\ ,\ r=\sqrt{z^2+{\left(\frac{\sqrt{2}}{2}L\right)}^2}\] 
\[{\vec{F}}_z=-4k\frac{Qq}{{\left(z^2+{\left(\frac{\sqrt{2}}{2}L\right)}^2\right)}^{\frac{3}{2}}}z\ \hat{k}\] 

Same as previous problem, first we must calculate each of the electric forces due to the $2\,\mu{\rm C}$ , $-4\,{\rm C}$ charges exerts on the third charge then use the superposition principle to determine the net electric force on it.
\begin{align*} 
{\vec{F}}_{21}&=k\frac{\left|q_1q_2\right|}{r^2_{12}}{\hat{r}}_{21}\\
&=9\times {10}^9\frac{2\times {10}^{-6}\times 7\times {10}^{-6}}{{\left(0.5\right)}^2}{\hat{r}}_{21}\\
&=0.504\ {\hat{r}}_{21}\,{\rm N}
\end{align*}
${\hat{r}}_{21}$ is the unit vector points from $q_2$ toward $q_1$ so if one decomposes it, we get
\[{\vec{F}}_{21}=0.504\left(\frac{1}{2}\hat{x}+\frac{\sqrt{3}}{2}\hat{y}\right){\rm N}\] 
(Notation: $F_{12}$ is the force exerted by point charge $q_1$on point charge $q_2$)
\begin{align*}
{\vec{F}}_{31}&=k\frac{\left|q_1q_3\right|}{r^2_{13}}{\hat{r}}_{31}\\
&=9\times {10}^9\frac{\left|7\times {10}^{-6}\times (-4)\times {10}^{-6}\right|}{{\left(0.5\right)}^2}\left({\cos  60{}^\circ \ }\hat{x}+{\sin  60{}^\circ \ }\left(-\hat{y}\right)\right)\ {\rm N}
\end{align*}
\[{\vec{F}}_{31}=1.008\ \left(\frac{1}{2}\hat{x}+\frac{\sqrt{3}}{2}\left(-\hat{y}\right)\right)\ {\rm N}\] 
Using superposition principle: ${\vec{{\rm F}}}_{{\rm 1}}{\rm =}{\vec{{\rm F}}}_{{\rm 31}}{\rm +}{\vec{{\rm F}}}_{{\rm 32}}$, we obtain
\begin{align*}
{\vec{{\rm F}}}_{{\rm 1}}&=0.504\left(\frac{1}{2}\hat{x}+\frac{\sqrt{3}}{2}\hat{y}\right)+1.008\ \left(\frac{1}{2}\hat{x}+\frac{\sqrt{3}}{2}\left(-\hat{y}\right)\right)\\
&=0.756\ \hat{x}-0.437\ \hat{y}\ \ ({\rm N)}
\end{align*}
\[\left|{\vec{F}}_1\right|=\sqrt{{\left(0.756\right)}^2+{\left(-0.437\right)}^2}=0.873\,{\rm N}\] 
And the direction of the resultant force with the horizontal axis ($x$) is 
\[\alpha={{\tan }^{-1} \left(\frac{\left|-0.437\right|}{\left|0.756\right|}\right)\ }=30.02{}^\circ \ \] 
Since $F_{1x}>0\ $and $F_{1y}<0$ so the net force lies in the fourth quadrant.

(a) Note: electric field defined as the electric force per unit charge. i.e. $\vec{{\rm E}}=\frac{\vec{{\rm F}}}{q}$. 
The net electric field of a system of point charges at any point obeys the superposition principle. 
At the location of $q$, because of the $+3q$, there is an electric field
\[{\vec{E}}_3=k\frac{3q}{a^2+a^2}\hat{r}=\frac{3}{2}k\frac{q}{a^2}\hat{r}\] 
Where we have used the Pythagoras theorem $r^2=a^2+a^2$. $\hat{r}$ is the unit vector points from $3q$ toward $q$ that must be decomposed in $\hat{x}$ and $\hat{y}$ directions, as follows
\[\hat{r}=\hat{x}\,{\cos  \theta\ }+\hat{y}\,{\sin \theta\ }\] 
So ${\vec{E}}_3=E_{3x}\hat{x}+E_{3y}\hat{y}=\frac{3}{2}k\frac{q}{a^2}\left({\cos \theta\ }\hat{x}+{\sin \theta\ }\hat{y}\right)$ but ${\tan \theta\ }=\frac{a}{a}=1\Rightarrow \theta=45{}^\circ $
The electric fields due to the charges $+2q$ at the $+q$ location are: 
\[{\vec{E}}_2=k\frac{2q}{a^2}\ \hat{x}\ \ ,\ \ {\vec{E}}^{'}_2=k\frac{2q}{a^2}\hat{y}\] 
Therefore the net electric field of three point charges $+3q,+2q,+2q$ at the location of the fourth charge is (by using the superposition principle): 
\begin{align*}
{\vec{E}}_{net}&={\vec{E}}_3+{\vec{E}}_2+{\vec{E}}^{'}_2\\
&=\frac{3}{2}k\frac{q}{a^2}\left({\cos  45{}^\circ \ }\hat{x}+{\sin  45{}^\circ \ }\hat{y}\right)+k\frac{2q}{a^2}\left(\hat{x}+\hat{y}\right)\\
&=k\frac{q}{a^2}\left[\left(\frac{3}{2}{\cos  45{}^\circ \ }+2\right)\hat{x}+\left(\frac{3}{2}{\sin  45{}^\circ \ }+2\right)\hat{y}\right]\\
&=\left(2+\frac{3}{2\sqrt{2}}\right)k\frac{q}{a^2}(\hat{x}+\hat{y})
\end{align*}
But we know that the magnitude and direction of a vector with components $\vec{a}=a_x\hat{x}+a_y\hat{y}$ is
$\left|a\right|=\sqrt{a^2_x+a^2_y}$  and $\alpha={{\tan }^{-1} \left(\frac{\left|a_y\right|}{\left|a_x\right|}\right)\ }$ with positive $x$ axis. 
Therefore, the magnitude and direction of the net electric field is 
\[\left|{\vec{E}}_{net}\right|=\left(2+\frac{3}{2\sqrt{2}}\right)k\frac{q}{a^2}\ \sqrt{2}=k\frac{q}{a^2}\left(2\sqrt{2}+\frac{3}{2}\right)\ \] 
\[\alpha={{\tan }^{-1} 1\ }=45{}^\circ !\] 
(b) From the definition of the electric field ${\vec{{\rm F}}}_{{\rm net}}{\rm =}q{\vec{{\rm E}}}_{{\rm net}}$. So 
\[{\vec{{\rm F}}}_{{\rm net}}=\left(2+\frac{3}{2\sqrt{2}}\right)k\frac{q^2}{a^2}(\hat{x}+\hat{y})\] 
And $\left|{\vec{{\rm F}}}_{{\rm net}}\right|=k\frac{q^2}{a^2}\left(2\sqrt{2}+\frac{3}{2}\right)$
 

Note: the electric force vector between two point charges located at distance $r$ from each other is $\vec{F}=k\frac{\left|q_1q_2\right|}{r^2}\hat{r}$. 
When there is a system of point charges and we want to find the net force on one of the charges, we must use the superposition principle i.e. the vector sum of the individual electric forces on the desired charge: ${\vec{{\rm F}}}_{{\rm net}}{\rm =}{\vec{{\rm F}}}_{{\rm 1}}{\rm +}{\vec{{\rm F}}}_{{\rm 2}}{\rm +}{\vec{{\rm F}}}_{{\rm 3}}+\dots $
So we must vector sum the individual forces of four point charges on the $-q$ in the center. 
The drawing below shows the direction of the individual forces. 
\[\left|F_1\right|=\left|F_2\right|=k\frac{\left|\left(-q\right)\left(+q\right)\right|}{{\left(0.3\right)}^2}=+\frac{kq^2}{0.09}\] 
\[\left|F_3\right|=\left|F_4\right|=k\frac{\left|\left(-q\right)\left(-q\right)\right|}{{\left(0.3\right)}^2}=+k\frac{q^2}{0.09}\] 
Because the $F_1,F_2$ and $F_3,F_4$ are separately in opposite directions to each other (i.e. ${\vec{F}}_1=-{\vec{F}}_2$ and ${\vec{F}}_3=-{\vec{F}}_4$) so the net force is
\[{\vec{F}}_{net}={\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3+{\vec{F}}_4=0\] 

(a) By definition, the total charge distributed in a region of space is $Q=\int{\rho\left(r\right)dV}$ 
\begin{align*}
Q_{tot}=\int{\rho\left(r\right)dV} &=\int^R_0{\rho_0\left(1-\frac{r}{R}\right)\ 4\pi r^2dr}\\
&= 4\pi \rho_0{\left(\frac{1}{3}r^3-\frac{r^4}{4R}\right)}^R_0\\
&=\frac{1}{3}\pi \rho_0R^3\\
&=\frac{1}{3}\pi R^3\left(\frac{3Q}{\pi R^3}\right)=Q\\
\ \ \ \Rightarrow Q_{tot}=Q
\end{align*}
(b) Use Gauss's law to find the electric field of a charge distribution inside and outside of the desired volume. The charge configuration has spherical symmetry so by symmetry consideration; the electric field must be radial.
\[\oint{\vec{E}.\hat{n}dA}=\frac{Q_{ins}}{\epsilon_0}\ \ ,\ \ {\rm Gauss}{{\rm s}}^{{\rm '}}{\rm s\ law}\] 
Where $Q_{ins}$ is the charge inside of a closed surface.
\[E_r\oint{dA}=\frac{Q_{ins}}{\epsilon_0}\to E_r\left(4\pi r^2\right)=\frac{Q_{ins}}{\epsilon_0}\Rightarrow E_r=\frac{Q_{ins}}{4\pi\epsilon_0r^2}\] 
if $r>R$ then $E_r=\frac{Q_{ins}}{4\pi\epsilon_0r^2}$ where $Q_{ins}=\int{\rho\left(r\right)dV}=Q$ because the Gaussian surface encloses the overall sphere. Therefore 
\[E_r\left(r>R\right)=\frac{Q}{4\pi\epsilon_0r^2}\ ,\ {\rm that\ is\ similar\ to\ th}{\rm e\ electric\ field\ of\ a\ point\ charge}\] 
(c) from part (b) , $E_r=\frac{Q_{ins}}{4\pi\epsilon_0r^2}$ but the amount of enclosed charge in a Gaussian surface inside of the sphere is 
\begin{align*}
Q_{ins} &=\int{\rho\left(r\right)dV}\\
&=\int^r_0{\rho_0\left(1-\frac{r}{R}\right)4\pi r^2dr}\\
&=4\pi\epsilon_0{\left(\frac{1}{3}r^3-\frac{r^4}{4R}\right)}^r_0\\
&=4\pi\epsilon_0\left(\frac{1}{3}r^3-\frac{r^4}{4R}\right)
\end{align*} 
\[\ \Rightarrow E_r\left(r<R\right)=\frac{\rho_0}{\epsilon_0}\left(\frac{1}{3}r-\frac{r^2}{4R}\right)\] 
(d) we must find the maximum value of the electric field inside of the sphere so 
\[\frac{d}{dr}E_r\left(r<R\right)=\frac{\rho_0}{\epsilon_0}\left(\frac{1}{3}-\frac{r}{2R}\right)=0\Rightarrow r=\frac{2}{3}R\] 
\[E_r\left(r=\frac{2}{3}R\right)=\frac{\rho_0}{\epsilon_0}\left(\frac{2}{9}R-\frac{R}{9}\right)=\frac{\rho_0}{9\epsilon_0}R\] 
 

Solution: you can find the solution of this problem here.

Solution: the solution is here. 

(a) The volume charge density is defined as $\rho=\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi R^3}\ $.
(b) First from Gauss's law determine the electric field outside the sphere:
\[\oint{\vec{E}.d\vec{A}}=\frac{Q_{en}}{\epsilon_0}\ \ \Longrightarrow \ E_{out}\left(4\pi r^2\right)=\frac{Q}{\epsilon_0}\to E_{out}=\frac{Q}{4\pi\epsilon_0r^2}\ \ like\ a\ point\ charge!\] 
Now use the definition of the potential difference as below:
\[V\left(r\right)-V\left(\infty \right)=-\int^r_{\infty }{\vec{E}\left(r\right).d\vec{r}}=-\int^r_{\infty }{\frac{Q}{4\pi \epsilon_0}\frac{dr}{r^2}}\] 
\[\therefore V\left(r>R\right)=\ \frac{Q}{4\pi \epsilon_0}\frac{1}{r}\ \ ,\ \ V\left(r=R\right)=\frac{Q}{4\pi \epsilon_0}\frac{1}{R}\] 
(c) Similar as previous, use Gauss's law inside the sphere to find the $E_{in}$: 
\[\oint{\vec{E}.d\vec{A}}=\frac{Q_{en}}{\epsilon_0}\ \Longrightarrow E\left(r\right)\left(4\pi r^2\right)=\frac{1}{\epsilon_0}\int^r_0{\rho dv}=\frac{1}{\epsilon_0}\rho\left(\frac{4}{3}\pi r^3\right)\ \] 
\[\Rightarrow E_{in}\left(r\right)=\frac{\rho}{3\epsilon_0}r\ \ or\ E_{in}\left(r\right)=\frac{Q}{4\pi \epsilon_0R^3}\ r\ \] 
\[V\left(r\right)-V\left(R\right)=-\int^r_R{{\vec{E}}_{in}\left(r\right).d\vec{r}}=-\frac{Q}{4\pi \epsilon_0R^3}\ \int^r_R{rdr}=-\frac{Q}{4\pi \epsilon_0R^3}\frac{1}{2}{\left.r^2\right|}^r_R\] 
\[V\left(r\right)-\frac{Q}{4\pi \epsilon_0R}=-\frac{Q}{4\pi \epsilon_0R^3}\frac{1}{2}\left(r^2-R^2\right)\ \] 
\[\therefore V\left(r<R\right)=\frac{Q}{8\pi\epsilon_0R}\left(3-{\left(\frac{r}{R}\right)}^2\right)\] 
(d)

(e)  
        (i) The charge density inside a conductor is always zero i.e. $\rho_{in}=0$
        (ii) $V\left(r>R\right)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$
        (iii) $V\left(r<R\right)=\frac{Q}{4\pi \epsilon_0}\frac{1}{R}$
 

(a) Due to cylindrical symmetry, the $\vec{E}$ field lines are cylindrically out.
For $r<R$: $\rho_0\int{\vec{E}}.\hat{n}da=Q_{en}=\int^r_0{\rho\left(r^{'}\right)dv}$
\[\epsilon_0E\left(2\pi rL\right)=\int^r_0{\rho_0\frac{r^{'}}{R}\ (r^{'}dr^{'}d\phi dz)}=\frac{\rho_0}{R}\ 2\pi L\frac{r^3}{3}\] 
\[\therefore \ \vec{E}\left(r<R\right)=E_{in}=\frac{\rho_0}{3\epsilon_0R}\ r^2\ \hat{r}\] 
$\hat{r}$ is the unit vector along the radius and $dv$ is the volume element of the cylinder. 
For $r>R$ : $$\epsilon_0\int{\vec{E}}.\hat{n}da=Q_{en}=\int^R_0{\rho\left(r^{'}\right)dv}=2\pi L\frac{\rho_0}{R}\frac{R^3}{3}$$
\[\therefore \ \vec{E}\left(r>R\right)=E_{out}=\frac{\rho_0}{3\epsilon_0r}\ R^2\ \hat{r}\] 
(b) The potential difference is defined by $V\left(r\right)-V\left(a\right)=-\int^r_a{\vec{E}.d\vec{l}}$
\[r>R\ :V\left(r\right)=-\int^r_a{\frac{\rho_0}{3\epsilon_0r^{'}}\ R^2}dr^{'}=-\frac{\rho_0R^2}{3\epsilon_0}{\ln  \left(\frac{r}{a}\right)\ }\] 
\begin{align*}
r<R:\ \ V\left(r\right) &=-\int^r_a{\vec{E}.d\vec{l}}=-\int^R_a{{\vec{E}}_{out}.d\vec{r}}-\int^r_R{{\vec{E}}_{in}.d\vec{r}}\\
&=-\int^R_a{\frac{\rho_0}{3\epsilon_0r}\ R^2dr}-\int^r_R{\frac{\rho_0}{3\epsilon_0R}\ r^2dr}\\
&=-\frac{\rho_0R^2}{3\epsilon_0}{\ln  \left(\frac{R}{a}\right)\ }-\frac{\rho_0}{3\epsilon_0R}{\left.\frac{r^3}{3}\right|}^r_R\\
&=-\frac{\rho_0R^2}{3\epsilon_0}{\ln  \left(\frac{R}{a}\right)\ }-\frac{\rho_0}{3\epsilon_0R}\frac{1}{3}\ (r^3-R^3)
\end{align*}  
(c) Infinity cannot be used as reference point, since the charge distribution extends to $\infty $. You can also this by the problem caused if we set $a\to \infty $.
 

(a) The potential difference between two arbitrary points is defined as below
\begin{align*}
\Delta V_{AB}&=-\int^B_A{\vec{E}.d\vec{S}}\\
&= -\int^B_A{12\left(-\hat{j}\right).\left(dx\hat{i}+dy\hat{j}\right)}\\
&=+12\int^B_A{dy}\\
&=+12\int^{y_B}_{y_A}{dy}\\
&=12\int^0_4{dy}\\
&=12{\left(y\right)}^0_4\\
&=12\left(0-4\right)=-48\ {\rm V}
\end{align*}  
In above, $d\vec{S}$ is the displacement vector between $A$ and $B$.
(b)  
\begin{align*}
\Delta V_{AC}&=-\int^A_C{\vec{E}.d\vec{S}}\\
&=-\int^A_C{12\left(-\hat{j}\right).\left(dx\ \hat{i}+dy\ \hat{j}\right)}\\
&=12\int^{y_B}_{y_A}{dy}\\
&=12\int^0_4{dy}=-48\ {\rm V}
\end{align*} 
\[\Delta V_{AB}=\Delta V_{AC}\] 
(c) The work done on a charge $q$ in an electric field is $W=q\Delta V$. Therefore, 
\[W=q\Delta V=\left(0.05\right)\left(-48\right)=-2.4\ {\rm J}\] 
The minus sign indicates that we must does external work to displace $q$ from $A$ to $C$.
 

(a) Use the definition of the electric field of a point charge to calculate its magnitude and direction at arbitrary distance.
\begin{align*}
\vec{E_1}&=k\frac{\left|q_1\right|}{r^2_1}\ {\hat{r}}_1\\
&=k\frac{\left|q_1\right|}{r^2_1}\ \left(\hat{j}\,{\cos \theta\ }-\hat{i}\,{\sin \theta\ }\right)\\
&=k\frac{\left|q_1\right|}{r^2_1}\ \left(\hat{j}\frac{1}{\sqrt{5}}-\hat{i}\frac{2}{\sqrt{5}}\right)\\
&=9\times {10}^9\frac{5\times {10}^{-6}}{0.05}\ \left(\hat{j}\frac{1}{\sqrt{5}}-\hat{i}\frac{2}{\sqrt{5}}\right) 
\end{align*}
So $\vec{E_1}=9\times {10}^5\left(\hat{j}\frac{1}{\sqrt{5}}-\hat{i}\frac{2}{\sqrt{5}}\right)=\left(+402.49\ \hat{j}-804.98\hat{i}\right)\times {10}^3\ \left(\frac{{\rm N}}{{\rm C}}\right)$
In above, from the geometry, we have 
\begin{gather*}
{\cos \theta\ }=\frac{10}{r_1}=\frac{10}{\sqrt{{10}^2+{20}^2}}=\frac{1}{\sqrt{5}}\\
{\sin \theta\ }=\frac{20}{r_1}=\frac{2}{\sqrt{5}}\\
r^2_1={\left(0.1\right)}^2+{\left(0.20\right)}^2=0.05
\end{gather*}
(b) Similar as part (a), 
\begin{align*}
\vec{E_2}&=k\frac{\left|q_2\right|}{r^2_2}{\hat{r}}_2\\
&=9\times {10}^9\frac{\left(10\times {10}^{-6}\right)}{{\left(0.2\right)}^2}\hat{i}\\
&=2250\times {10}^3\hat{i}
\end{align*}
(c) Using the superposition principle, the vector sum of each of the electric fields at arbitrary point $P$ is the total electric field at that point, we obtain
\[\vec{E_P}=\vec{E_1}+\vec{E_2}=\left(1445.02\hat{i}+402.49\hat{j}\right)\times {10}^3\ {\rm N/C}\] 
(d) Using the definition of electric field in general, we obtain
\begin{align*}
\vec{F_e}=e\vec{E_e}\ \Rightarrow \ \left|\vec{F_e}\right|&=\left|e\right|\left|\vec{E_P}\right|\\
&=(1.6\times {10}^{-19})\times {10}^3\sqrt{{\left(1445.02\right)}^2+{\left(402.49\right)}^2}\ {\rm N}
\end{align*}
\[\left|\vec{F_e}\right|{\rm \ \sim 2.4\times }{{\rm 10}}^{{\rm -}{\rm 13}}{\rm N}\] 

(a) The potential electric due to a point charge at distance $r$ is defined by: $V\left(r\right)=\frac{q}{4\pi \epsilon_0r}$. When there is a system of point charges, the potential electric of this system at an arbitrary point P is defined by the sum of each of the potential electric at that point. Therefore, 
\[V\left(O\right)=V_q+V_{-2q}+V_{3q}\] 
\[V\left(O\right)=\frac{q}{4\pi \epsilon_0}\left[\frac{1}{L}-\frac{2}{L}+\frac{3}{\sqrt{L^2+L^2}}\right]=\frac{q}{4\pi \epsilon_0L}\frac{3\sqrt{2}-2}{2}\ \] 

(b) The electric field of a system of point charges is the vector sum of each of the electric fields at that point (superposition Principle).
\[\vec{E}\left(O\right)={\vec{E}}_q+{\vec{E}}_{-2q}+{\vec{E}}_{3q}\] 
Note: the electric field of a point charge at distance $r$ is calculated by 
\[\vec{E}\left(r\right)=\frac{1}{4\pi \epsilon_0}\frac{\left|q\right|}{r^2}\hat{r}\] 
Where $\hat{r}$ is the unit vector along the line from the point charge toward the desired location $r$.
\[\vec{E}\left(O\right)=\frac{1}{4\pi \epsilon_0}\left[\frac{q}{L^2}\hat{i}-\frac{2q}{L^2}\hat{j}+\frac{3q}{2L^2}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right]=\frac{q}{4\pi \epsilon_0L^2}\left[\hat{i}\frac{4+3\sqrt{2}}{4}-\hat{j}\frac{8-3\sqrt{2}}{4}\right]\] 
 

First find the total electric field of the wire at point $P$ in terms of $Q$ and $L$. To do this, choose a charge element $dq=\lambda dx$ at distance $x$ from the origin. This element is similar to the point charge, thus one can use the electric field definition and find its magnitude at point $P$( see the figure), then integrate over all of the wire to compute the total field at that point as follows:
\begin{align*}
E=\int{dE}&=\int{\frac{kdq}{{\left(0.5-x\right)}^2}}\\
&=k\lambda\int^L_0{\frac{dx}{{\left(0.5-x\right)}^2}}\\
&=k\lambda{\left(\frac{1}{0.5-x}\right)}^L_0\\
&=k\lambda\left(\frac{1}{0.5-L}-\frac{1}{0.5}\right)\\
&=k\lambda\frac{L}{0.5(0.5-L)}
\end{align*}
Recall that the total charge of the wire is $Q=\lambda L$ so 
\[E=\frac{kQ}{0.5(0.5-L)}\] 
Now substitute the given values of $E=300\ {\rm N/C}$ and $Q=+3\ {\rm nC}$, solving for $L$, we obtain
\[E=\frac{kQ}{0.5\left(0.5-L\right)}\Rightarrow 300=\frac{\left(9\times {10}^9\right)\left(3\times {10}^{-9}\right)}{0.5\left(0.5-L\right)}\ \] 
\[\Rightarrow L=0.32\ {\rm m}\] 

The electric potential of a point charge at distance $r$ is given by $V\left(r\right)=kq/r$. In the case of continuous charges, we must firstly find the electric potential due to an infinitesimal charge element at distance $r$, then integrate over all of the charge configuration. Thus first calculate the electric potential of a uniformly charged ring at distance $z$ on the axis of the ring
\begin{align*}
V=\int{dV}&=\int{\frac{kdq}{r}}\\
&=k\int^{2\pi}_0{\frac{\lambda\,ad\phi}{\sqrt{a^2+z^2}}}\\
&=\frac{k\lambda a}{\sqrt{a^2+z^2}}(2\pi)
\end{align*}
Since we are asked that the electric potential must be computed at the center of the ring so setting $z=0$ we get 
\[V_{ring}\left(center\right)=\frac{kQ}{a}\] 
As we see, this electric potential at center of the ring is similar to electric potential of a point charge! We have calculated the total charge of the ring as $Q_{ring}=\lambda L=\left(2\pi a\right)\lambda$, where $L$ is the circumference of the ring.
Let $d$ be the distance of point charge $q$ from the center of the ring, so its electric potential at the center of the ring is 
\[V_{point}=\frac{kq}{d}\] 
The electric potential due to a system of point charges is the sum of the potentials due to each of these charges. Thus the total potential of this configuration at the center of the ring is  
\[V=\Sigma V_i=V_{point}+V_{ring}=\frac{kq}{d}+\frac{kQ}{a}\] 
Setting $V_{tot}=0$, we obtain
\[0=\frac{kq}{d}+\frac{kQ}{a}\Rightarrow d=-a\frac{q}{Q}=-16\left(-\frac{18}{24}\right)=12\ {\rm cm}\] 
 

First compute the force between these two point charges at the distance stated in the problem as follows
\[F=\frac{k\left|+eQ\right|}{r^2}\] 
Now using Newton's second law and solving for $Q$, we obtain
\[F=ma{\rm =}\frac{k\left|+eQ\right|}{r^2}\Rightarrow \left|Q\right|=\frac{mar^2}{k(+e)}\] 
By substituting the given values of the problem, one can find the value of charge $Q$, thus
\[\left|Q\right|=\frac{\left(1.67\times {10}^{-27}\right)\left(4.31\times {10}^{26}\right){\left(8\times {10}^{-14}\right)}^2}{\left(9\times {10}^9\right)\left(1.6\times {10}^{-19}\right)}=3.06\times {10}^{-18}{\rm C}\] 
Let $d$ be the closest distance to the center of the nucleus, at this point proton coming to a stop that is its kinetic energy becomes zero so use conservation of mechanical energy between these points to find the desired distance.
\[E_i=E_f\] 
\[\underbrace{\frac{k(+e)Q}{r}}_{U_i}+\underbrace{\frac{1}{2}mv^2_i}_{K_i}=\underbrace{\frac{k(+e)Q}{d}}_{U_f}+\underbrace{\frac{1}{2}mv^2_f}_{K_f=0}\] 
Rearranging and solving for $d$, we obtain
\[d=\frac{k(+e)Q}{\frac{1}{2}mv^2_i-\frac{keQ}{r}}\] 
\[\Rightarrow d=\frac{\left(9\times {10}^9\right)\left(1.67\times {10}^{-19}\right)\left(3.06\times {10}^{-18}\right)}{\frac{1}{2}\left(1.67\times {10}^{-27}\right)\left(4.15\times {10}^6\right)-\frac{\left(9\times {10}^9\right)\left(1.67\times {10}^{-19}\right)\left(3.06\times {10}^{-18}\right)}{8.00\times {10}^{-14}}}\] 

\[\Rightarrow d\sim \ 6.4\times {10}^{-14}\ {\rm m}\] 
Note: the electric potential energy of a point charge $q$ that is located at distance $r$ from an  another charge $Q$ is given by $U=kqQ/r$. 
 

First calculate the electric field of an infinitesimal element of the line at the point $P$, then by integrating over all of the length find the total electric field of the arc at $P$. Thus
\[{\vec{E}}_{tot}=\int{d\vec{E}}=\int{\frac{kdq}{R^2}\hat{r}}=\frac{k\lambda R}{R^2}\int{d\theta\ \hat{r}}\] 
Recall that the uniformly charge distribution and line element is related together by $dq=\lambda dl$, where in the case of arc $dl=ds=Rd\theta$. $\hat{r}$ is the unit vector along the direction of the total electric field and must be decomposed into  $\hat{i}$ and $\hat{j}$. Therefore,
\begin{align*}
{\vec{E}}_{tot}&=\frac{k\lambda}{R}\int^{\frac{\pi}{2}}_0{\left({\cos \theta\ }\,\hat{i}+{\sin \theta\ }\,\left(-\hat{j}\right)\right)d\theta}\\
&=\frac{k\lambda}{R}{\left({\sin \theta\ }\,\hat{i}+{\cos \theta\ }\,\hat{j}\right)}^{\frac{\pi}{2}}_0\\
&=\frac{k\lambda}{R}\left(\hat{i}-\hat{j}\right)
\end{align*}
This is the total electric field at point $P$. Its direction is explicitly shown in the equation and figure. Before finding the magnitude of it, we must explicitly express the total electric field in terms of the charge $q$ of the arc. By using the definition of the linear charge density $\lambda$, we get
\[\lambda=\frac{q}{L}=\frac{q}{R\left(\frac{\pi}{2}\right)}\] 
Where, in this case, $L$ is the circumference of the arc.  Thus, substituting the $\lambda$ into ${\vec{E}}_{tot}$, we obtain
\[{\vec{E}}_{tot}=\frac{k}{R}\frac{q}{R\left(\frac{p}{2}\right)}\left(\hat{i}-\hat{j}\right)\to {\vec{E}}_{tot}=\frac{2kq}{\pi R^2}\left(\hat{i}-\hat{j}\right)\] 
The magnitude of the total electric field is computed as follows
\[\left|{\vec{E}}_{tot}\right|=\sqrt{E^2_x+E^2_y}=\frac{2\sqrt{2}}{\pi}\frac{kq}{R^2}\ \] 
And its direction is 
\[\theta={{\tan }^{-1} \left|\frac{E_y}{E_x}\right|\ }={{\tan }^{-1} \left|\frac{1}{-1}\right|\ }=45{}^\circ \] 
Since $E_x>0\ $and $E_y<0$, then the total electric field lie in the 4${}^{th}$ quadrant. 
 

Calculate the electric field of the rings at distance $d$ on the $z$ axis then use the superposition principle to find the total electric field at that point. 
Choose the $z$ axis to coincide with the axis of the ring with the ring in the $z=0$ plane. The electric field of a ring at a point on its axis through the center of the ring is calculated as follows
\[{\vec{E}}_z=\int{dE_z}=\int{\frac{k\overbrace{dq}^{\lambda Rd\phi }}{r^2}{\cos \theta\ }\hat{k}}=k\lambda R\int^{2\pi}_0{\frac{{\cos \theta\ }}{r^2}d\phi}\] 
Where $ds=Rd\phi$ is the length of the arc in the polar coordinate. By symmetry consideration, the $x$ and $y$ components of the electric field due to an element of charge $dq$ is nonzero i.e. $E_x=E_y=0$. As can be seen from the geometry 
\[{\cos \theta\ }=\frac{d}{r}\ \ and \ \ r=\sqrt{R^2+d^2}\] 
Therefore, 
\begin{align*}
{\vec{E}}_z=\int{dE_z}&=\int^{2\pi}_0{\frac{k\lambda R}{r^2}\frac{d}{r}\ d\phi\ \hat{k}}\\ 
&=\frac{k\lambda Rd}{{\left(R^2+d^2\right)}^{\frac{3}{2}}}\underbrace{\int^{2\pi}_0{d\phi}}_{2\pi}\\
&=\frac{kQd}{{\left(R^2+d^2\right)}^{\frac{3}{2}}}\hat{k}
\end{align*}
Where we have used the definition of the linear charge density on the ring 
\[\lambda=\frac{Q}{L}=\frac{Q}{2\pi R}\] 
Where $L$ is the circumference of the ring. 
Now using the superposition principle and setting ${\vec{E}}_{tot}=0$ and solving for $d$, we obtain the desired location
\[{\vec{E}}_{tot}={\vec{E}}_R+{\vec{E}}_{2R}=\frac{kQd}{{\left(R^2+d^2\right)}^{\frac{3}{2}}}\hat{k}+\frac{k(-\sqrt{8}Q)d}{{\left({\left(2R\right)}^2+d^2\right)}^{\frac{3}{2}}}\hat{k}=0\] 
\[\Rightarrow {\left(4R^2+d^2\right)}^{\frac{3}{2}}=\underbrace{\sqrt{8}}_{2^{\frac{3}{2}}}{\left(R^2+d^2\right)}^{\frac{3}{2}}\to {\left(4R^2+d^2-2\left(R^2+d^2\right)\right)}^{\frac{3}{2}}=0\] 
The argument of the equation above must be zero to satisfy the condition, thus 
\[4R^2+d^2-2\left(R^2+d^2\right)=0\Rightarrow 2R^2-d^2=0\] 
\[\therefore d=\sqrt{2}R\] 
 

Solution 1: By definition, the electric flux passing through any surface with area element $dA$ is the scalar product of the normal component of the electric field and area of the surface that is 
\[{\Phi }_E=\int{\vec{E}.\hat{n}dA}\] 
Where $\hat{n}$ is the unit vector normal to the surface. 
In the case of hemisphere or sphere the unit vector is along the radius i.e. $\hat{n}=\hat{r}$. Since $\vec{E}$ is parallel to the axis of hemisphere the scalar product of $E$ and $\hat{r}$ is $E{\cos \theta\ }$. The range of polar angle is $\theta=0$ to $\theta=\pi/2\ $.Therefore,
\begin{align*} {\Phi }_E=\int^{\frac{\pi}{2}}_0{\vec{E}.\hat{r}dA} &= \int^{2\pi}_0{d\phi}\int^{\frac{\pi}{2}}_0{E{\cos \theta\ }\ R^2{\sin \theta\ }d\theta}\\ &= 2\pi ER^2\int^{\frac{\pi}{2}}_0{\underbrace{{\sin \theta\ }{\cos \theta\ }}_{\frac{1}{2}{\sin  2\theta\ }}d\theta} \end{align*} 
\begin{align*} \Rightarrow {\Phi }_E &=2\pi ER^2\left(\frac{1}{2}\right){\left(-\frac{1}{2}{\cos  2\theta\ }\right)}^{\frac{\pi}{2}}_0\\ &=-\frac{1}{2}\pi ER^2\left({\cos  2\frac{\pi}{2}\ }-{\cos  0\ }\right)\\ &=\pi ER^2 \end{align*}
In above, $dA=R^2{\sin \theta\ }d\theta d\phi $ is the area element of the sphere. 
Solution 2: all of the electric lines through the circle at the bottom of the hemisphere, passing through the area of the hemisphere. So by calculating the flux through the circle, one can find the flux through the hemisphere but the important thing to remember is that, by convention, the flux through the circle is incoming and is negative of the outgoing flux through the hemisphere that is ${\Phi }_E\left(circle\right)=-{\Phi }_E(hemisphere)$. Therefore, by definition of the electric flux through a surface is
\[{\Phi }_E\left(circle\right)=\vec{E}.\hat{n}\ A=E\left(+\hat{k}\right).\left(-\hat{k}\right)\left(\pi R^2\right)=-\pi ER^2\] 
Note: the normal vector $\hat{n}$ is always out of one side of a surface (as in the above case).

First we must determine the magnitude of the electric field of a uniformly charged sphere at its surface. To do this, using Gauss's law we obtain the electric field inside of it
\[\oint_S{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}\] 
Where the symbol $\oint $ represents the integral over a closed surface and $Q_{enc}$ is stands for the charge enclosed by the closed surface $S$. To compute the electric field inside of such a sphere, one must consider an imaginary spherical surface of radius $r$ , called Gaussian surface, then evaluate the integral above as follows
\[\oint_S{{\vec{E}}_{in}.\hat{n}dA}=E_{in}\oint_s{dA}=\frac{Q_{enc}}{\epsilon_0}\] 
We have taken $E_{in}$ out of the integral because it is constant everywhere on the surface. The integral of $dA$ over the spherical surface is just the total area of it that is $4\pi r^2$. By definition of the volume charge density, we can find the amount of charge enclosed by this imaginary sphere. 
\[Q_{enc}=\rho V_{enc}=\frac{4}{3}\pi r^3 \rho\] 
using this, the electric field inside of a uniformly charged sphere with charge density $?$ is 
\[E_{in}\left(4\pi r^2\right)=\frac{4}{3}\pi r^3\frac{\rho}{\epsilon_0}\Rightarrow E_{in}=\frac{\rho}{3\epsilon_0}r\] 
Since $\rho$ is constant, we can use this fact to express $E_{in}$ in terms of the total charge of sphere $Q$. 
\[\rho=\frac{Q}{V_{tot}}=\frac{Q}{\frac{4}{3}\pi R^3}\] 
By substituting $\rho$ into $E_{in}\left(r\right)$, we get
\[E_{in}\left(r\right)=\frac{1}{3\epsilon_0}\frac{Q}{\frac{4}{3}\rho R^3}r=\frac{Q}{4\pi \epsilon_0}\frac{r}{R^3}\] 
Setting $r=R$, we obtain the electric field of a uniformly charged sphere at its surface
\[{\vec{E}}_{in}\left(r=R\right)=\frac{Q}{4\pi \epsilon_0}\frac{1}{R^2}(\hat{i})\] 
Now compute the electric field due to the positive point charge $q$ at point $P$ as follows
\[{\vec{E}}_q=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}(-\hat{i})=\frac{1}{4\pi \epsilon_0}\frac{q}{R^2}(-\hat{i})\] 
By applying the superposition principle, we can find the net electric field of above configuration at point $P$
\[{\vec{E}}_{net}={\vec{E}}_{in}\left(r=R\right)+{\vec{E}}_q=\frac{Q}{4\pi \epsilon_0}\frac{1}{R^2}(\hat{i})+\frac{1}{4\pi \epsilon_0}\frac{q}{R^2}\left(-\hat{i}\right)\] 
Setting $E_{net}=0$, we can determine the unknown charge $q$ 
\[E_{net}=0\Rightarrow \frac{Q}{4\pi \epsilon_0}\frac{1}{R^2}-\frac{1}{4\pi \epsilon_0}\frac{q}{R^2}=0\ \] 

\[\Rightarrow \ q=Q\ \] 
In above, we have used the convention that direction of the electric field of a positive point charge is outward and vice versa. 

The charge $-Q$ induces charge $+Q$ on the inner sphere and similarly this charge induces charge $-Q$ on the outer sphere of radius $2R$. Since the electric field inside a conductor in electrostatic equilibrium is zero and since it is stated that the shell has a net charge $+2Q$, we conclude that the excess charge must be reside on the outer surface of the shell. So by summing these two contribution over the outer surface, one can find the total free charge uniformly distributed on the outer surface to be
\[Q^{'}=+2Q+\left(-Q\right)=+Q\] 
 

(a) In such a problem, one can use the Gauss's law to find the electric field everywhere. Gauss's law states that the electric flux through any closed surface $S$ is equal to the charged enclosed by it divided by $\epsilon_0$. The mathematical form of it is 
\[\oint_s{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}\] 
To use the Gauss's law, we must firstly consider a closed surface which is called Gaussian surface. This surface has the same symmetry as the electric field. In this case, the Gaussian surface must be a spherical surface of radius $r$ concentric with the conducting charged sphere of radius $a$. 

Note: although the Gauss's law is true for any surface surrounding a charged configuration, but it is useful only when we choose a Gaussian surface to match the original symmetry of the problem. 

To find the electric field in region $r<a$, the Gaussian surface is inside the sphere of radius $a$. Since inside this surface does not any enclosed charge, so there is no flux through it, so
\[\oint_s{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}=0\Rightarrow E\left(r<a\right)=0\] 
In the region $a<r<b$, the Gaussian surface encloses the charged sphere $q$, so 
\[\oint_s{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}=\frac{q}{\epsilon_0}\Rightarrow E\left(a<r<b\right)\oint_s{dA}=\frac{q}{\epsilon_0}\] 
\[\Rightarrow \ E\left(a<r<b\right)=\frac{q}{4\pi \epsilon_0r^2}\] 
Since $E$ is constant and perpendicular everywhere on the Gaussian surface, so we have taken out of integral. The closed integral $\oint_s{dA}$ is the surface area of the sphere. In the case of sphere, the normal vector $\hat{n}$ is along the radial direction i.e. $\hat{n}=\hat{r}$. Since inside the Gaussian surface there is a positive charge then the electric field point away from the center or is radially outward i.e. $\vec{E}=E\left(r\right)\hat{r}$. 

Region $b<r<c$ lies inside the conductor. Using this fact the electric field inside a conductor is zero , so $E\left(b<r<c\right)=0$. In region $r>c$, the net charge encloses by the Gaussian surface is $+q$, so in this region the electric field is 
\[\oint_s{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}=\frac{q}{\epsilon_0}\Rightarrow E\left(r>c\right)=\frac{q}{4\pi \epsilon_0r^2}\] 

(b) The graph is as follows

(c) The free charge $q$ inside the sphere of radius $a$ induces the charge $-q$ on the inner surface and subsequently this charge also induces the charge $+q$ on the outer surface of the spherical shell. 
 

The magnitude of the electric force between two point charges $q$ and $q^{'}$ located at distance $r$ from each other is given by the Coulomb's law as follows 
\[F=k\frac{\left|qq^{'}\right|}{r^2}\] 
Where $k=9\times {10}^9\ {\rm N.}{{\rm m}}^{{\rm 2}}/C^2$. The directions of the forces the two charges exert on each other are always along the line joining them. 
The figure below is a free body diagram for proton. Let us consider the charge $q$ to be positive. In such a case, $F$ is the force exerted on the proton by the electron and $F^{'}$ is the force exerted by the charge $q$ on it. Now compute these forces and use the superposition principle to find the total force acting on the proton.
\[{\vec{F}}_{tot}=\vec{F}+{\vec{F}}^{'}=k\frac{\left|\left(-e\right)(+e)\right|}{{{x^{'}}^2}}\left(-\hat{i}\right)+k\frac{\left|\left(+e\right)q\right|}{{\left(x-x^{'}\right)}^2}\left(+\hat{i}\right)\] 
Since the charge $q$ is in equilibrium state so the total force exerted on it must be zero
\[{\vec{F}}_{tot}=0\ ,{\rm \ equilibrium\ condition}\] 
\[k\frac{\left|\left(-e\right)\left(+e\right)\right|}{{x^{'}}^2}\left(-\hat{i}\right)+k\frac{\left|\left(+e\right)q\right|}{{\left(x-x^{'}\right)}^2}\left(+\hat{i}\right)=0\] 
\[\Rightarrow \frac{e}{{x^{'}}^2}=\frac{\left|q\right|}{{\left(x-x^{'}\right)}^2}\Rightarrow e{\left(x-x^{'}\right)}^2=\left|q\right|{x^{'}}^2\] 
\[e{\left(4\ {\rm nm-1nm}\right)}^{{\rm 2}}=\left|q\right|{\left(1{\rm nm}\right)}^{{\rm 2}}\] 
\[\Rightarrow \ \left|q\right|=9e\] 
If we assume that the charge $q$ is negative, we get the same result.

The electric force acting on the charge $q$ is $\vec{F}=q\vec{E}=q\left|\vec{E}\right|(-\hat{j})$, so using Newton's second law the magnitude of acceleration in the $y$ direction is 
\[a_y=\frac{F}{m}=\frac{q\left|\vec{E}\right|}{m}\] 
The initial velocity and height of the droplet is $v_0=0\ ,\ y_0=0$, respectively. Thus by applying the following kinematical relation $y=\frac{1}{2}at^2+v_0t+y_0$, we can obtain the distance traveled in $y$ direction in terms of the elapsed time $t$. 
\[y=\frac{1}{2}\left(\frac{q\left|\vec{E}\right|}{m}\right)t^2\] 
The elapsed time in $x$ and $y$ directions is the same. There are no forces in the $x$ direction so that the velocity $v_x=v_0$ is constant. Using the equation of uniform motion $x=vt$, we can find the elapsed time $t$ as follows
\[d=v_0t\Rightarrow t=\frac{d}{v_0}\] 
By substituting $t$ into the $y\ $equation above, we get
\[y=\frac{1}{2}\left(\frac{q\left|\vec{E}\right|}{m}\right){\left(\frac{d}{v_0}\right)}^2=\frac{q\left|E\right|d^2}{2mv^2_0}\] 
 

(a) Use Gauss's law to find the electric field due to a long uniform line of charge with charge density $\lambda$. In this case, the Gaussian surface must be cylindrical to match the symmetry of the problem so the Gaussian surface is a coaxial cylinder with the wire with length $L$ and radius $R$. The direction of $\vec{E}$ is also radial. Therefore,
\[\oint_S{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}\to E_R\oint_S{dA}=\frac{1}{\epsilon_0}\int^L_0{\lambda dx}\] 
\[\Rightarrow E_R\left(2\pi RL\right)=\frac{1}{\epsilon_0}\lambda L\Rightarrow E_R=+\frac{\lambda}{2\pi\epsilon_0R}\] 
In above, $\oint_s{dA}$ is the area of the curved surface of the cylinder. Since the $\vec{E}$ is radially outward from the wire, the contributions due to left $\hat{n}=-\hat{x}$ and right $\hat{n}=+\hat{x}$ end of cylinder are zero i.e. $\vec{E}.\hat{n}=0$. $R$ is the distance from the wire. 

Therefore, the line $+\lambda$ ,which is lies along the $x$ axis, at distance $y$ from it produced the electric field 
\[{\vec{E}}_1=\frac{+\lambda}{2\pi\epsilon_0y}\ \hat{y}\] 
And the line $+\lambda$, which is lies along the $y$ axis, at distance $x$ from it produce the following electric field
\[{\vec{E}}_2=\frac{+\lambda}{2\pi\epsilon_0x}\hat{x}\] 
Thus, the total electric field at a point in the $x-y$ plane is 
\[{\vec{E}}_{tot}=\frac{+\lambda}{2\pi\epsilon_0}\left(\frac{1}{y}\ \hat{y}+\frac{1}{x}\ \hat{x}\right)\] 

(b) The electric potential between $a$ and $b$ is defined as $V_b-V_a=-\int^b_a{\vec{E}.d\vec{l}}$. we are moving perpendicular to the wire 1 and subsequently parallel to the wire $2$, so wire

$2$ does not contribute to the potential difference. In this case, $d\vec{l}=dy\hat{y}$.
\[\Delta V=-\int^{3a}_a{{\vec{E}}_1.d\vec{l}}=-\int^{3a}_a{\frac{\lambda}{2\pi\epsilon_0y}dy}=-\frac{\lambda}{2\pi\epsilon_0}{\left.{\ln  y\ }\right|}^{3a}_a\] 
\[\therefore \ \Delta V=-\frac{\lambda}{2\pi\epsilon_0}{\ln  3\ }\] 
So going from $y=a$ to $y=3a$ the potential is decreasing since we are moving away from the charged line $1$. 

(c) In this case, we are approaching perpendicularly to the wire $2$. In this direction we have change in the electric field of ${\vec{E}}_2$, so only this wire contribute to the potential difference. 
\[\Delta V=-\int^{3a}_a{E_2\hat{x}.dx\ \hat{x}}=-\int^{3a}_a{\frac{\lambda}{2\pi\epsilon_0x}dx}=-\frac{\lambda}{2\pi\epsilon_0}{\left.{\ln  x\ }\right|}^{3a}_a\] 
\[\therefore \Delta V=-\frac{\lambda}{2\pi\epsilon_0}{\ln  3\ }\] 
(d) The work done on a point charge $q$ to move it from point $a$ to $b$ in an external electric field with a potential difference $\Delta V$ is found by $W=q\Delta V$. 
To going from $(3a,3a)$ to $\left(a,a\right)$ we can consider the following path 

\[\left(3a,3a\right)\xrightarrow{1}\left(3a,a\right)\xrightarrow{2}(a,a)\] 
In the path 1, we have 
\[\Delta V_1=V_{(3a,3a)}-V_{(3a,a)}=-\int^{3a}_a{{\vec{E}}_1\hat{y}.dy\left(-\hat{y}\right)}=\frac{\lambda}{2\pi\epsilon_0}{\ln  3\ }\] 
In the path 2, we also have
\[\Delta V_2=V_{3a,a}-V_{a,a}=-\int^{3a}_a{E_2\hat{x}.dx\left(-\hat{x}\right)}=\frac{\lambda}{2\pi\epsilon_0}{\ln  3\ }\ \] 
The signs in the above potentials changes because we are moving in the opposite direction as was considered in previous parts. Therefore, the total electric potential between the $(3a,3a)$ and $(a,a)$ is 
\[\Delta V=\Delta V_1+\Delta V_2=\frac{\lambda}{\pi\epsilon_0}{\ln  3\ }\] 
The work done on the charge $q$ by an external agent is found as
\[W=-q\Delta V=-\frac{q\lambda}{2\pi\epsilon_0}{\ln  3\ }\] 
The electric fields due to the line of charges does positive work on the external charge $q$.

(e) Since in this case, unlike the spherical charged distribution, the charge is extended to the infinity and thus at that point we have charge and electric field!
 

Once the two spheres are connected, they constitute an equipotential system. In this case, some of the charge $Q$ delivers to the sphere $2$, that is the total charge of the system becomes $Q=Q_1+Q_2$ where $Q_1$ is the new charge on the sphere $1$ and $Q_2$ is one the sphere $2$. Recall that the electric potential due to a sphere of radius $R_1$ and charge $Q_1$ is $V_1=kQ_1/R_1$, and for a sphere of radius $R_2$ with charge $Q_2$ is $V_2=kQ_2/R_2$. 

Since after the two spheres connect to each other form an equipotential system, we have
\[V_1=V_1\Rightarrow \frac{kQ_1}{R_1}=\frac{kQ_2}{R_2}\Rightarrow Q_1=\frac{R_1}{R_2}Q_2\] 
Substituting $Q_1$ into the $Q=Q_1+Q_2$ and solving for $Q_2$, we obtain
\[Q=\frac{R_1}{R_2}Q_2+Q_2=Q_2\left(1+\frac{R_1}{R_2}\right)\] 
\[\Rightarrow Q_2=\frac{QR_2}{R_2+R_1}\] 
 

The potential energy of two point charges separated by distance $r$ is 
\[U=k\frac{q_1q_2}{r}\] 
The potential energy is a scalar quantity so the total potential energy due to a configuration of point charges is the sum of the potential energy of each pairs of them. In this case, since the distance of charges and their magnitudes are the same so 
\begin{align*}
U&=U\left(3\ electrons\ pairs\right)+U\left(3\ (e^++e^-)\ pairs\right)\\
&=3k\frac{\left(-e\right)\left(-e\right)}{L}+3k\frac{\left(+e\right)\left(-e\right)}{r}
\end{align*}
$r$ is the distance between electron and proton which can be computed from the geometry below
\begin{align*}
\therefore U_{tot}&=+3k\frac{e^2}{L}-\frac{3ke^2}{\frac{L}{2{\cos  30{}^\circ \ }}}\\
&=\frac{3ke^2}{L}\left(1-2{\cos  30{}^\circ \ }\right)\\
&=\frac{3\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{5.1\times {10}^{-9}}\left(1-2\, \cos 30{}^\circ \right)\\
&=-9.92\times {10}^{-20}\ {\rm J}
\end{align*}
From the definition of cosine of an angle in the triangle above, we get
\[{\cos  30{}^\circ \ }=\frac{\frac{L}{2}}{r}\Rightarrow r=\frac{L}{2{\cos  30{}^\circ \ }}\] 
 

Conservation of mechanical energy is conserved, $\Delta K+\Delta U=0$. Recall that the potential energy of a point charge moving through a potential difference $\Delta V$ is $\Delta U=q\Delta V$. Since the electron accelerated from rest; hence $\Delta K=K_f-K_i=\frac{1}{2}mv^2_f-0$. A negative charge moves against the electric field which is the direction of the decreasing potential energy i.e. $\Delta V=V_f-V_i<0$. Thus
\[\Delta K+\Delta U=0\] 
\[\frac{1}{2}m\left(v^2_f-v^2_0\right)+q\Delta V=0\Rightarrow v_f=\sqrt{-\frac{2q\Delta V}{m}}\] 
Electron has $q=-e=-1.6\times {10}^{-19}\ {\rm C}$
\[\Rightarrow v_f=\sqrt{\frac{2\left(1.6\times {10}^{-19}\right)\left(9900\right)}{9.31\times {10}^{-31}\ {\rm kg}}}=5.83\times {10}^7\frac{{\rm m}}{{\rm s}}\] 
 

The electric field $\vec{E}$ and potential difference $V$ are related by 
\[\Delta V=V_f-V_i=-\int^f_i{\vec{E}.d\vec{s}}\] 
Where $s$ is the displacement vector from point $i$ to point $f$. Note that only the component of the electric field parallel to the line of integration is relevant. Therefore,
\[\Delta V=V\left(r=8{\rm m}\right){\rm -}V\left(r={\rm 6m}\right){\rm =-}\int^{r=8}_{r=6}{\frac{1}{r^2}\hat{r}.dr\hat{r}}={\rm -}\int^{r=8}_{r=6}{\frac{1}{r^2}dr}\] 
\[{\rm =-}{\left(-\frac{1}{r}\right)}^8_6=\left(\frac{1}{8}-\frac{1}{6}\right)=\frac{3-4}{24}=-\frac{1}{24}\ {\rm V}\] 
In above, $d\vec{s}=dr\ \hat{r}$ is chosen in the radial direction. 
 

An electron in a conductor will propagate randomly or have thermal motion similar to the motion of the molecules of the gas such as air. With applying an electric field, we can change this randomly motions into a uniform motion in one direction with speed $v_d$ which is called drift speed. The typical value for such a speed is about ${10}^{-4}\ {\rm m/s}$. 
The relation between current through a wire with cross sectional area $A$ and drift speed of charge carries, the charges that move in a conductor, of charge density $n$ is
\[I=nqAv_d\] 
In this figure, the current through $3\ {\rm mm}$ surface is the same as that of $1\ {\rm mm}$ surface that is $I_3=I_1$, therefore
\begin{align*}
A_3v_{d3}=A_1v_{d1}\Rightarrow v_{d1}=\frac{A_3}{A_1}v_{d3}&={\left(\frac{r_3}{r_1}\right)}^2v_{d3}\\
&={\left(\frac{3}{1}\right)}^2\left(4\times {10}^{-4}\right)\\
&=36\times {10}^{-4}\frac{{\rm m}}{{\rm s}}{\rm \ }
\end{align*}
$A=\pi r^2$is the cross sectional area of the circular disk. 
 

The surface of a conductor is an equipotential surface which means that whose surface must be always in equilibrium. Putting an excess charge would disrupt this equilibrium condition on the surface since causes an electric field forms between the accumulation of charges in that point and other points of the surface. This electric field, by definition, causes a change in the electric potential difference between any points on the surface, instead, if the excess charge spreads uniformly over the surface the equipotential condition of the surface is not violated. 

The correct answer is E.
 

the electric field of a point charge is found as $E=kQ/r^2$, which obeys the inverse-square law but the electric field of an electric dipole, pairs of point charges with equal magnitude and opposite sign separated by a distance $d$, at far away distances from the dipole can be found as
\[{\vec{E}}_{dipole}=\frac{1}{4\pi {\epsilon }_0}\frac{\vec{p}}{r^3}\] 
Where $\vec{p}=q\vec{d}$ is the electric dipole moment and $\vec{d}$ is distance vector from the negative charge to the positive charge. 
As you can see the electric field of a dipole decreases more rapidly than that of point charges. 

The correct answer is B. 
 

recall that positive charge moves in direction of electric field and a negative charge moves in opposite direction of electric field. Since the displacement of electron, to north, and direction of the electric field, to north, are in the same direction, the electron is decreasing down. 

The correct answer is B.
 

The amount of electric flux  ${\mathrm{\Phi }}_E$ through any closed surface and the associated enclosed total charge is related together by Gauss's law as 
\[{\mathrm{\Phi }}_E=\frac{Q_{encl}}{{\epsilon }_0}\] 
Hemisphere has two surfaces, rounded and flat base thus the total electric flux through it is 
\[{\mathrm{\Phi }}_{E.r}+{\mathrm{\Phi }}_{E.b}=\frac{Q_{encl}}{{\epsilon }_0}\] 
\[9.8\times {10}^4+{\mathrm{\Phi }}_{E.b}=\frac{6.6\times {10}^{-7}}{8.854\times {10}^{-12}}\Rightarrow \ {\mathrm{\Phi }}_{E.b}=-2.34\ \mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}\] 

The correct answer is C.
 

If we have symmetric configuration, as this case, use Gauss's law to find the electric field at each point of space. To do this, one must suppose a Gaussian surface at the desired point. In this problem, draw a Gaussian surface as sphere of radius of $R/2$ and proceed as follows
\[\oint{\vec{E}.d\vec{A}}=\frac{Q_{encl}}{{\epsilon }_0}\] 
Where $Q_{encl}$ is the charge enclosed in the Gaussian surface which here is sphere of radius $R/2$ and is determined by the definition of volume charge density 
\[\rho =\frac{Q}{V}\to Q_{encl}=\rho V_{Gauss}=\rho \left(\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\right)=\frac{1}{6}\rho \pi R^3\] 
Substituting above into Gauss's law, we get
\[E\oint{dA}=\frac{1}{6}\rho \pi R^3\] 
The electric field is taken out of the integral since it is constant at the location of the Gaussian sphere by symmetry considerations. The closed integral gives the area of the Gaussian surface. Therefore
\[E\left(4\pi {\left(\frac{R}{2}\right)}^2\right)=\frac{\frac{1}{6}\rho \pi R^3}{{\epsilon }_0}\to \ \ E=\frac{1}{4\pi {\epsilon }_0}\frac{2}{3}\rho \pi R\] 
Since the density of the sphere is uniform everywhere so substitute it by the total charge induced on the original sphere of radius $R$ as 
\[\rho =\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi R^3}\] 
Therefore
\[E=\frac{1}{4\pi {\epsilon }_0}\frac{2}{3}\frac{Q}{\frac{4}{3}\pi R^3}\pi R=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{{2R}^2}=\frac{Q}{8\pi {\epsilon }_0R^2}\] 

The correct answer is D.
 

The potential difference between two points is defined as the ratio of the change in electric potential energy to the electric charge which is moved between those points and is written as \[\Delta V=\frac{\Delta U}{q}\] 
We know that the work done on a system is stored in its potential energy so, regardless of its sign, we have $W=\Delta U=500\quad \rm J$. Hence, \[\Delta V=\frac{500}{40}=12.5\quad \rm V\] To practice more solved problems on the electric potential, refer here.

The correct answer is A.

The electric field between two large parallel conducting plates is uniform. Recall that the electric field of a positive plane is away from it and a negative plane is toward it, using this hint and the superposition principle determine the electric field between the plate as follows

In above we have used the definition of the uniform electric field in terms of electric potential as $E=V/d$, since the distance between the plates is the same so the correct answer is D.
 

The electric potential energy of a pair of point charges can be found as follows 
\[\mathrm U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1q_2}{r}\] 
Therefore,
\[U=\left(9\times {10}^9\right)\frac{\left(5.5\times {10}^{-6}\right)\left(-2.3\times {10}^{-8}\right)}{\left(3.5\times {10}^{-2}\mathrm{m}\right)}=-3.3\times {10}^{-2}\mathrm{J}\] 
When the point charges are placed at infinity their potential energy is zero!

The correct answer is B.
 

The electric potential between the two point a and b is defined as
\[V_b-V_a=-\int^b_a{\vec{E}.d\vec{s}}\] 
Where $\vec{s}$ is the displacement vector from a toward b and is parallel to the field line. The negative sign indicates that the electric potential at point b is lower than at point b. 
To find the electric potential it is customary to choose the reference of the electric potential to be $V=0$ at $r_a=\infty $. Therefore find the electric potential between $r_a=\infty $ to $r_b=0$. The electric field inside and outside a conducting shell is $E_{in}=0$ and $E_{out}=kQ/r^2$, where $Q\ $is the total charge on the sphere. 
\[V_0-V_{\infty }=-\int^0_{\infty }{\vec{E}.d\vec{s}}=-\int^R_{\infty }{E_{out}dr}-E_{in}\int^0_R{dr}\] 
\[V_0-0=-\int^R_{\infty }{\frac{kQ}{r^2}dr}-0\int^0_R{dr}={{\left.\frac{kQ}{r}\right|}^R}_{\infty }=\frac{kQ}{R}\] 
\[\Rightarrow V_0=\frac{kQ}{R}=V\] 
$kQ/R$ is the electric potential of a conducting charged sphere at its surface.

The correct answer is a. 

In the first equality, $\vec{E}.d\vec{s}$ substituted by $Edr$ since the electric field outside is radial and the displacement vector $d\vec{s}$ has been also chosen radially.