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Three points charges are located on a circular arc as shown in the figure.

(a) What is the total electric field at P, the center of the arc?

(b) Find the electric force that would exerted on an electron with charge $q=-e$ placed at P.

(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,

\[{\vec{E}}_1=\frac{k\left|q_1\right|}{r^2_1}{\hat{r}}_1=\frac{kQ}{a^2}\left({\cos 60{}^\circ \ }\hat{i}+{\sin 60{}^\circ \ }\left(-\hat{j}\right)\right)\]

\[{\vec{E}}_2=\frac{k\left|q_2\right|}{r^2_2}{\hat{r}}_2=\frac{k\left(2Q\right)}{a^2}\left(-\hat{i}\right)\]

\[{\vec{E}}_3=\frac{k\left|q_3\right|}{r^2_3}{\hat{r}}_3=\frac{kQ}{a^2}\left({\cos 60{}^\circ \ }\hat{i}+{\sin 60{}^\circ \ }\hat{j}\right)\]

\[{\vec{E}}_P={\vec{E}}_1+{\vec{E}}_2+{\vec{E}}_3=\frac{kQ}{a^2}2\left({\cos 60{}^\circ \ }-1\right)\hat{i}=-\frac{kQ}{a^2}\hat{i}\]

(b) The force exerted on a charge $q$ in an 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 positive charge.

In the figure below, let $Q_1=+6.84\, \mu{\rm C}\ ,\ Q_2=-16.2\,\mu {\rm C}$ and $Q_3=+9.2\, \mu {\rm C}$ be rigidly fixed and separated by $r_{12}=r_{13}=r_{23}=1.34\, {\rm m}$. Calculate the electric potential at point P which is half-way between $Q_1$ and $Q_2$.

Two protons are near the surface of the earth. The first proton is placed securely on the ground. How high will the second proton have to be placed in order to just remain stationary above the Earth's surface?

Two equal but oppositely charges spheres ($\pm 5\times {10}^{-8}{\rm C}$) are hanging vertically from $10\,{\rm cm}$ strings. These spheres each have mass of $2\, {\rm g}$, and are separated by a very thin insulating sheet. At $t=0$, an external electric field is applied horizontally, and the charges separate, coming to equilibrium when an angle of $20{}^\circ $ is between the two strings.

(a) Is the external electric field pointing to the right or to the left?

(b) What is the magnitude of the external electric field?

Consider a sphere with uniform charge density $\rho$ Suppose that at a point ${\mathbf a}$ from the origin a spherical cavity, free of charge is made. The spherical cavity is entirely within the sphere. Calculate the electric field vector inside the cavity (hint: use the concept of superposition).

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.

\[{\vec{E}}_{+\rho}(P)=+\frac{\rho}{3\epsilon_0}\vec{r}\]

\[{\vec{E}}_{-\rho}(P)=\frac{-\rho}{3\epsilon_0}(\vec{r}-\vec{a})\]

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

\[\Rightarrow {\vec{E}}_P=+\frac{\rho}{3\epsilon_0}\vec{a}\]

If the electric potential in a region is given by $V\left(x\right)=6/x^2$, what is the $x$ component of the electric field in that region?

A $+4.0\, \mu {\rm C}$ point charge and a $-4.0\, \mu {\rm C}$ point charges are placed as shown in the figure. What is the potential difference $V_A-V_B$ between points A and B?

Find the potential difference $(V_B-V_A)$ between point $A(x=0,y=0)$ and point $B(x=5{\rm m,y=-5m)}$ for an electric field $\vec{E}=500\hat{i}-200\hat{j}\ {\rm (}\frac{{\rm V}}{{\rm m}}{\rm )}$.

A $200\ {\rm m}$ long thin wire carries a line charge density $\lambda=264\ {\rm nC/m}$. find the potential difference between points $5.0\, {\rm m}$ and $6.0\, {\rm m}$ on a perpendicular radius to the axis of the wire, provided the perpendicular radius is not near either end of the wire.

Three point charges are fixed in place in the right triangle shown below, in which $q_1=0.71\, \mu {\rm C}$ and $q_2=-0.67\, \mu {\rm C}$. What is the magnitude and direction of the electric force on the $+1.0\, \mu {\rm C\ }$(let's call this $q_3$) charge due to the other two charges?

\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)}{{\left(0.10{\rm cm}\right)}^2}\left({\cos \theta\ }\hat{x}+{\sin \theta\ }(-\hat{y})\right)\ {\rm N}

\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\left(0.8\ \hat{x}+0.6\left(-\hat{y}\right)\right)\\

&=\left(0.511\hat{x}-0.383\hat{y}\right){\rm \ 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}\ {{\rm cm}}^{{\rm 2}}}\left(-\hat{y}\right)\\

&=\left(-1.675\ \hat{y}\right)\ {\rm N}

\end{align*}

Therefore, the resultant force on the $q_3$ is

\[{\vec{F}}_3={\vec{F}}_{13}+{\vec{F}}_{23}=0.511\hat{x}-0.383\hat{y}+\left(-1.675\ \hat{y}\right)=\left(0.511\ \hat{x}-2.058\ \hat{y}\right)\ {\rm N}\]

The direction of the net force with the $x$ axis are determined by ${\tan \alpha\ }=\left|F_y\right|/\left|F_x\right|$, 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 $\left|{\vec{F}}_3\right|=\sqrt{{\left(0.511\right)}^2+{\left(-2.058\right)}^2}=2.12\ {\rm N}$

Two small insulating spheres are attached to silk threads and aligned vertically as shown in the figure. These spheres have equal masses of $40\, {\rm g}$, and carry charges $q_1$ and $q_2$ of equal magnitude $2.0\, \mu {\rm C}$ but opposite sign. The spheres are brought into the positions shown in the figure, with a vertical separation of $15\, {\rm cm}$ between them. Note that you cannot neglect gravity. What is the tension in the lower threads?

A point charge $Q=-500\ {\rm nC}$ and two unknown point charges $q_1$ and $q_2$ are placed as shown in the figure. The electric field at the origin, due to charges $Q,q_1$ and $q_2$ is equal to zero. What is the amount of the charge $q_1$?

The electric field strength in the space between two closely spaced parallel disks is $1.0\times {10}^5\ {\rm N/C}$. This field is the result of transferring $3.9\times {10}^9$ electrons from one disk to the other. What is the diameter of the disks?

At a distance of $5.8\, {\rm cm}$ from the center of a very long uniformly charged wire, the electric field has magnitude $2000\, {\rm N/C\ }$and is directed toward the wire. What is the charge on a $1.00\, {\rm cm}$ length of wire near the center?

In the figure, charge $q_1=+2.5\,{\rm nC}$ is located at the origin, charge $q_2=2.0\,{\rm nC}$ is located on the $x$-axis at $x=+4.0{\rm cm}$, and point P is located at $x=+4.0\,{\rm cm},\ y=+3.0\,{\rm cm}$.

(a) Calculate the magnitude and direction of the electric field at point P.

(b) Calculate the electric potential at point P.

(c) A $+2.0\,{\rm nC}$ charge is placed at point P. What is the magnitude of the electric force this charge experiences?

\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{\left|q_1\right|}{r^2_1}{\hat{r}}_1\\

&=k\frac{2.5\times {10}^{-9}}{{\left(0.04\right)}^2+{\left(0.03\right)}^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\ \left(-\hat{y}\right)\\

&={10}^{-9}k\left(8\hat{x}-16.22\hat{y}\right)

\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\frac{{\rm N}}{{\rm C}}

\end{align*}

Where we have substituted the $k=8.99\times {10}^9\frac{{\rm N.}{{\rm m}}^{{\rm 2}}}{C^2}$

(b) The electric potential due to a point charge $q$ at distance $r$ is defined as $V\left(r\right)=k\frac{q}{r}$

\[V_1\left(P\right)=k\frac{q_1}{r_1}=k\frac{-2\times {10}^{-9}}{0.03}=-599.3\ {\rm V}\]

\[V_2\left(P\right)=k\frac{q_2}{r_2}=\left(8.99\times {10}^9\right)\frac{2.5\times {10}^{-9}}{0.05}=449.5\ {\rm V}\]

\[\therefore V_P=V_1+V_2=-149.8\ {\rm V}\]

(c) If a charge is located in a uniform electric field $E$, then a force with magnitude $F=qE$ exert on it.

\[F_P=qE_P=\left(2\times {10}^{-9}\right)\left(162.58\right)=325.16\ {\rm nN}\]

An electron was accelerated from rest through a potential difference of $1300\ {\rm V}$. What is its speed?

A doubly ionized alpha particle ($q=+2e,\ m=4m_p$) is moving in a uniform parallel electric field in a direction opposite of the electric field. When measurements commence, it has a kinetic energy of $56.4\,{\rm MeV}$ and it continuously decelerate to a stop. If it moves $6620\, {\rm km}$ during this time, what is strength of the electric field?

A particle with unknown mass $m$ and charge $q$ is initially moving at constant velocity $v=4\times {10}^5\ {\rm m/s}$ in a region with zero electric field. It then enters a region where the field is $E=100\ {\rm N/C}$, in the same direction as the particle's motion. The particle travels for a distance of $x=1.2\ {\rm cm}$ before momentarily coming to a stop. What is the particle's charge to mass ratio $q/m$? Be sure to include the sign of the ratio and its units.

An electron (mass = $9.11\times {10}^{-31}{\rm kg}$) is released from rest in a uniform electric field of magnitude of $5000\ {\rm N/C}$. How long would it take the electron to reach a speed of $100\frac{{\rm m}}{{\rm s}}$?

A line of positive charge is formed into semicircle of radius $R$. The charge per unit length along the semicircle is described by the expression $\lambda=\lambda_0\,{\cos \theta\ }$. Find:

(a) The total charge on the semicircle.

(b) The electric field vector at point P

\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*}

A uniformly charged insulating rod of length $14\,{\rm cm}$ is bent into the shape of a semicircle. The rod has a total charge of $-7.5\, \mu{\rm C}$. Find the magnitude and the direction of the electric field at $O$, the center of the semicircle. Calculate the force on a $3\,\mu{\rm C}$ charge placed at the point P.

Three charged particles are aligned along the $y$ axis. Find the electric field at $P$ point. Determine the electric force on a $Q$ charged place on the $P$ point.

A small $2{\rm g}$ plastic ball is suspended by a $20\ {\rm cm}$ long string in a uniform electric field as shown in the figure. If the ball is in equilibrium when the string makes a $15{}^\circ $ angle with the vertical, what is the net charge on the ball?

Four identical particles, each having charge $+q$, are fixed at the corners of a square of side $L$. A fifth point charge $-Q$ (at $P$ point) lies a distance $z$ along the line perpendicular to the plane of the square and passing through the center of the square. Determine the force exerted by the other four charges on $-Q$.

Three point charges are located at the corners of an equilateral triangle an in the figure. Find the magnitude and direction of the net electric force on the $7\, \mu{\rm C}$ charge.

Four point charges are at the corners of a square of side $a$ as in the figure.

(a) Determine the magnitude and direction of the electric field at the location of charge $q$

(b) What is the resultant force on $q$.

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 $\left|{\vec{{\rm F}}}_{{\rm net}}\right|=k\frac{q^2}{a^2}\left(2\sqrt{2}+\frac{3}{2}\right)$

Four point charges are at the corners of a square. The distance from each corner to the center is $0.3\ {\rm m}$. At the center there is a $-q$ point charges. What is the magnitude of the net force on this charge?

A non-uniform, but spherically symmetric, distribution of charge has a charge density of $\rho\left(r\right)=\rho_0\left(1-\frac{r}{R}\right)$ for $(r<R)$ and $\rho\left(r\right)=0$ for $(r>R)$ where $\rho_0=\frac{3Q}{\pi R^3}$ is a positive constant.

(a) Show that the total charge contained in the charge distribution is $Q$

(b) Show that the electric field in the region ($r>R$) is identical to that produced by a point charge $Q$ at $r=0$.

(c) Obtain an expression for the electric field for $r<R$ and graph the magnitude of the $E(r)$ as a function of $r$.

(d) Find the value of $r$ at which the electric field $E(r)$ is a maximum, and find the value of that maximum field.

A non-uniform electric field is given by the expression

\[\vec{E}=ay\hat{x}+bz\hat{y}+cx\hat{z}\]

Where $a,b,c$ are constants, $\hat{x},\hat{y},\hat{z}$ are unit vectors in the $x,y,z$ directions, respectively. Determine the electric flux through a rectangular surface in the $xy$ plane, extending from $x=0$ to $x=w$ and from $y=0$ to $y=h$.

The cubical surface of side length $L=12\ {\rm cm}$ is shown in the electric field $\vec{E}=\left(950\,y\ \hat{i}+650\,z\hat{k}\right)\ {\rm V/m}$. Find the electric flux through the top face of the cube.

An insulating solid sphere of radius $R$ has a uniform positive charge $Q$.

(a) Find charge density $\rho$

(b) Find the electric potential at a point $r$ outside the sphere ($r>R$). Take the potential to be zero at $r=\infty $.

(c) Find the potential at a point inside the sphere ($r<R$)

(d) Plot potential $V(r)$ as a function of the distance from the center.

(e) What can you say about the results (a),(b) and (c) if we have conducting sphere having the same amount of charge on it?

An infinitely long non-conducting cylinder of radius $R$ carries a nonuniform volume charge density $\rho\left(r\right)=\rho_0\frac{r}{R}$, where $\rho_0$ is known constant and $r$ is the cylindrical radial distance from its axis.

(a) Using Gauss's law determine the electric field vector everywhere (i.e. $r<R$ and $r>R$)

(b) Determine the corresponding electric potential everywhere, choosing $V\left(a\right)=0$, i.e. potential reference chosen at $r=a$ where $a>R$.

(c) Can you choose infinity as potential reference? Give your reasoning.

There is a uniform electric field $\vec{E}=12\frac{V}{m}\ (-\hat{j})$ along the $-y$ axis and these points are given with coordinates $(0,4{\rm m)}$,$B\left(0,0\right),\ C(2{\rm m,0)}$. Find the potential differences:

(a) $\Delta V_{AB}$

(b) $\Delta V_{AC}$

(c) If a charge $q_0=0.05\ {\rm C}$ is taken from $A$ to $C$, find the work done on the charge.

The charge $q_1=-5\, \mu{\rm C}$ and $q_2=10\, \mu {\rm C}$ are located as in the figure.

(a) Find the electric field due to $q_1$ at the point $P$ located at $20\ {\rm cm}$ along $x$ axis.

(b) Find the electric field of $q_2$ at $P$

(c) What is the total electric field at $P$?

(d) If an electron ($q_e=-1.6\times {10}^{-19}\,C$) is put at $P$ what will be the magnitude of the electric force acting on the electron

Suppose $k=\frac{1}{4\pi \epsilon_0}=9\times {10}^9\ {\rm N.}{{\rm m}}^{{\rm 2}}{\rm /}{{\rm C}}^{{\rm 2}}$

Three different point charges are arranged at the corners of a square of side $L$ shown in the figure.

(a) What is the potential at the forth corner(taken as the origin, point $O$)

(b) What is the electric field __vector__ at point $O$? (Do not forget the unit vectors, simplify your expressions)

In the figure below, the thin straight wire of length $L$ has a uniformly distributed charge of $Q=+3.0\ {\rm nC}$. Its electric field at point $P$ has magnitude $E=300\ {\rm N/C}$. Find $L$.

\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

\[E=\frac{kQ}{0.5\left(0.5-L\right)}\Rightarrow 300=\frac{\left(9\times {10}^9\right)\left(3\times {10}^{-9}\right)}{0.5\left(0.5-L\right)}\ \]

\[\Rightarrow L=0.32\ {\rm m}\]

A uniformly charged thin ring has radius $a=16.0\, {\rm cm}$ and total charge $Q=+24\, \mu {\rm C}$. Where must a point charge $q=-18\, \mu {\rm C}$ be placed on the axis of the ring for the electric potential at the center of the ring to be zero?

A proton moves directly towards a stationary nucleus of charge $Q$. The proton has a speed of $4.15\times {10}^6\ {\rm m/s}$ and an acceleration of $4.31\times {10}^{26}\ {\rm m/}{{\rm s}}^{{\rm 2}}$ when it is $8.00\times {10}^{-14}\ {\rm m}$ from the center of the nucleus. What is the distance of the closest approach (to the center of the nucleus)? Mass of proton $m_{+e}=1.67\times {10}^{-27}{\rm kg}$ , $e=1.6\times {10}^{-19}{\rm C}$

A charge $q$ is distributed uniformly on a quarter side circle of radius $R$ as shown in the figure. What is the magnitude of the electric field at the center(the point $P$)?

Two concentric rings, one of radius $R$ and total charge $Q$ and the second of radius 2R and total charge $-\sqrt{8}Q$, are in the plane $z\ =\ 0$. The charge on each ring is distributed uniformly. Where on the positive $z$-axis is the electric field zero?

A hemispherical shell of radius $R$ is placed in an electric field $\vec{E}$ which is parallel to its axis. What is the flux ${\Phi }_E$ of the electric field through the shell?

A sphere of radius $R$ contains a total charge $Q$ which is uniformly distributed throughout its volume. At a distance $2R$ from the center of the sphere we place a point charge $q$. What is the value of $q$ which makes the electric field at the point $P$ on the surface zero?

A point charge of charge $-Q$ is placed at the center of a solid spherical conducting shell of inner radius $R$ and outer radius $2R$. The shell is in static equilibrium and has a net charge $+2Q$. What is the total charge on the outer surface (at $r=2R$) of the shell?

A solid conducting sphere carrying charge $q$ has radius $a$. It is inside a concentric hollow conducting sphere with inner radius $b$ and outer radius $c$. The hollow sphere has no net charge.

(a) Derive expressions for the electric field magnitude in terms of the distance $r$ from the center for the region $r<a\ ,\ a<r<b\ ,\ b<r<c\ $and $r>c$.

(b) Graph the magnitude of the electric field as a function of $r$ from $r=0$ to $r=2c$.

(c) What is the charge on the inner surface and on the outer surface of the hollow sphere?

A electron is fixed at the position $x=0$, and a second charge $q$ is fixed at $x=4\times {10}^{-9}\ {\rm m}$ (to the right). A proton is now placed between the two at $x^{'}=1\times {10}^{-9}\ {\rm m}$. What must the charge $q$ be (magnitude and sign) so that the proton is in equilibrium?

An ink droplet of mass $m$ and charge $q$ is injected horizontally with an initially velocity ${\vec{v}}_0$ into a region with an electric field $\vec{E}$ which is perpendicular to ${\vec{v}}_0$. If a piece of paper is positioned a distance $d$ away from the injection point as shown in the figure, what is the vertical deflection $y$ of the droplet?

A very long, uniform line of charge with positive linear charge density $+\lambda$ lies along the $x$ axis. An identical line of charge lies along the $y$ axis.

(a) Determine the electric field $\vec{E}(x,y)$ for all points in the $x-y$ plane.

(b) Determine the change in the electrostatic potential $\Delta V$ between the points $x=a,y=a$ and $x=a,y=3a$.

(c) Determine $\Delta V$ between the points $x=a,y=a$ and $x=3a,y=a$.

(d) How much work must be done to move a small negative charge $-q$ from the point $x=3a,y=3a$ to the point $x=a,y=a$?

(e) For a very long linear charge distribution, we do not define the zero of electrostatic potential to be an infinity. Why not?

A charge $Q$ is placed on a metal sphere (sphere $1$) of radius $R_1$ . Very far from this sphere is a second sphere (sphere $2$) of radius $R_2$ which is initially uncharged. If the two spheres are connected by a metal wire, what is the final charge $Q_2$ on sphere $2$?

Three electrons are placed at the vertexes of an equilateral triangle with side length of ${\rm 5.1\ nm}$. A proton is placed at the center of the triangle. What is the potential energy of this arrangement of charges?

An electron was accelerated from rest through a potential difference of ${\rm 9900\ V}$. What is its speed? ($m_e=9.31\times {10}^{-31}{\rm kg}$).

An electric field $\vec{E}=\frac{1}{r^2}\ \hat{r}$ is located in the $xy$ plane. What is the potential difference between $r=6\ {\rm m}$ and $r=8\ {\rm m}$?

\[\Delta V=V_f-V_i=-\int^f_i{\vec{E}.d\vec{s}}\]

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.

The figure shows a current entering a truncated solid cone made of a conducting metal. The electron drift speed at the ${\rm 3.0\ mm}$ diameter end of the cone is ${\rm 4.0\times }{{\rm 10}}^{{\rm -}{\rm 4}}{\rm \ m/s}$. What is the electron drift speed at the ${\rm 1.0\ mm}$ diameter end of the wire?

If excess charge is put on a spherical conductor,

(a) it remains where it was placed

(b) it spreads a little from where it was placed but not over the whole sphere

(c) it spreads uniformly over the surface of the sphere if the sphere is small

(d) it spreads uniformly throughout the volume of the conductor

(e) it spreads uniformly over the surface of the sphere

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.

Comparing the field of a single point charge with the field of an electric dipole,

(a) the field of the point charge decreases more rapidly with distance

(b) the field of the point charge decreases less rapidly with distance

(c) the field of the point charge decreases more rapidly with distance but only along the dipole axis

(d) the field of the point charge decreases less rapidly with distance but only perpendicular to the dipole axis

(e) the fields decrease equally rapidly with distance

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.

An electron traveling north enters a region where the electric field is uniform and points north. The electron:

(a) speeds up

(b) slows down

(c) veers east

(d) veers west

(e) continues with the same speed in the same direction

A $\mathrm{3.5\ cm}$ radius hemisphere contains a total charge of $6.6\times {10}^{-7}\mathrm{C}$. The flux through the rounded portion of the surface is $9.8\times {10}^4\mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}$. The flux through the flat base is:

(a) $0\ \mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}$

(b) $2.3\ \mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}$

(c) $-2.3\ \mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}$

(d) $-9.8\ \mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}$

(e) $9.8\ \mathrm{N.}{\mathrm{m}}^{\mathrm{2}}\mathrm{/C}$

Charge $Q$ is distributed uniformly throughout an insulating sphere of radius $R$. The magnitude of the electric field at a point $R/2$ from the center is:

(a) $Q/4\pi {\epsilon }_0R^2$

(b) $Q/\pi {\epsilon }_0R^2$

(c) $3Q/4\pi {\epsilon }_0R^2$

(d) $Q/8\pi {\epsilon }_0R^2$

(e) None of these

If $\mathrm{500\ J}$ of work are required to carry a $\mathrm{40\ C}$ charge from one point to another, the potential difference between these two points is:

(a) $12.5\, \mathrm V$

(b) $20000\, \mathrm V$

(c) $0.08\, \mathrm V$

(d) Depends on the path

(e) None of these

The diagram shows four pairs of large parallel conducting plates. The value of the electric potential is given for each plate. Rank the pairs according to the magnitude of the electric field between the plates, least to greatest

(a) 1,2,3,4

(b) 4,3,2,1

(c) 2,3,1,4

(d) 2,4,1,3

(e) 3,2,4,1

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.

A particle with a charge of $5.5\times {10}^{-6}\mathrm{C}$is $\mathrm{3.5\ cm}$ from a particle with a charge of $-\mathrm{2.3\times }{\mathrm{10}}^{\mathrm{-}\mathrm{8}}\mathrm{C}$. The potential energy of this two-particle system, relative to the potential energy at infinite separation, is:

(a) $3.3\times {10}^{-2}\mathrm{J}$

(b) $-3.3\times {10}^{-2}\mathrm{J}$

(c) $9.3\times {10}^{-1}\mathrm{J}$

(d) $-9.3\times {10}^{-1}\mathrm{J}$

(e) $0\ \mathrm{J}$

A hollow metal sphere is charged to a potential $V$. The potential at its center is:

(a) $V$

(b) 0

(c) $-V$

(d) $2\ \mathrm{V}$

(e) $\pi \ \mathrm{V}$

Category : electrostatic

Most useful formula in Electrostatic:

Coulomb's law: the force between two charged particle $q_1$ and $q_2$ separated by distance $r$ is

\[F=k\frac{|q_1q_2|}{r^2}\]

where

\[k=\frac{1}{4\pi \epsilon_0}=8.988 \times {10}^{9} \ {\rm N.m^{2}/C^{2}}\]

Electric field:

\[\vec E=\frac{\vec F}{q_0}\]

Electric field of a point charge at distance $r$:

\[\vec E=\frac{1}{4 \pi \epsilon_0}\frac{q}{r^2} \hat{r}\]

Gauss's law:

\[\Phi_B=\int{\vec E \cdot nd\vec A}=\frac{Q_{enclosed}}{\epsilon_0}\]

Number Of Questions : 61

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