# Electric Field - Problems and Solution

Problems and solutions on electric fields are presented for high school and college students. More practice problems in electric fields are also provided here

Electric and magnetic fields are vector quantities in physics. These quantities are described both with a magnitude and a direction (angle).

To have a better understanding of these quantities and their properties, refer to the page below:
Vector, definitions, formula, and solved problems

## Electric Field Practice Problems

Problem (1): What is the electric field due to a point charge of $20\,{\rm \mu C}$ at a distance of 1 meter away from it?

Solution: electric field formula at an arbitrary point from a point charge $q$ is \begin{align*} E&=k\frac{q}{r^2}\\ \\ &=9\times 10^{9}\frac{\left(20\times 10^{-6}\right)}{1^2}\\ \\&=1.8\times 10^{5} \quad {\rm \frac NC}\end{align*} where in above $k$ is an experimentally obtained positive constant called Coulomb constant whose exact value is $k=8.99\times 10^{9} \quad {\rm \frac{N\cdot m^2}{C^2}}$

Problem (2): In the vicinity of point charge $q$, we place a $0.2\,{\rm \mu C}$-charge so that a force of $5\times 10^{-5}\,{\rm N}$ applied on it due to the charge $q$. Find the electric field produced by this unknown charge $q$?

Solution: the Coulomb force exerted on a test point charge $q_0$ at any point is related to the electric field (due to an unknown charge $q$) at that point by $\vec{F}=q_0\vec{E}$Therefore, the magnitude of the electric field is obtained as \begin{align*}E&=\frac{F}{q_0}\\ \\ &={\rm \frac{5\times 10^{-5}\,N}{0.2\times 10^{-6}\,C}}\\ \\&= 250\quad {\rm \frac NC}\end{align*}

The electric field problems are a closely related topic to Coulomb's force problems

Problem(3): An electron is released from rest in a uniform electric field of magnitude $E=100\,{\rm N/C}$ and gains speed.

(a) Find the magnitude of the force applied to it?

(b) After traveling a distance of $1$ meter, how fast does it reach?

Solution: The magnitude of charge of electron is $e=1.6\times 10^{-19}\,{\rm C}$ and the electron mass is $m_e=9.1\times 10^{-31}\,{\rm kg}$. From rest means that $v_0=0$.

(a) The electric field and electric force is related by formula $F=qE$. Thus, we have \begin{align*}F&=qE\\&=(100)(1.6\times 10^{-19})\\&=1.6\times 10^{-17}\quad {\rm N}\end{align*}

(b) This part is a kinematics problem. "From rest" means the initial velocity of electron is zero, so $v_0=0$. By combining Newton's second law $F=ma$, and kinematic equation $v^{2}-v_0^{2}=2ax$ as below, we can obtain the speed of the electron after traveling one meter through the electric field. \begin{align*} v^{2}&=2\left(\frac{F}{m}\right)x+v_0^{2}\\ \\ &=2\left(\frac{1.6\times 10^{-17}}{9.1\times 10^{-31}}\right)\times (1)\\ \\ &=0.35\times 10^{14}\end{align*} Taking square root from both sides, we get $v\cong 6\times 10^{6}\quad {\rm m/s}$

Problem (4): What is the force on an electron at a point where the electric field is $\vec{E}=5\times 10^{5}\,(\hat{i})\,{\rm N/C}$.

Solution: electric force $\vec{F}$ on a test point charge $q_0$ and electric field $\vec{E}$ is related by $\vec{F}=q_0 \vec{E}$. Thus, the electric force is \begin{align*} \vec{F}&=-e \vec{E}\\ &=-\left(1.6\times 10^{-19}\right)\left(5\times 10^{5}\,\hat{i}\right)\\&=-8\times 10^{-14}\,\hat{i}\quad {\rm N}\end{align*} Where $q_0$ is replaced by the electron charge which has a negative value i.e. $q_0=e=-1.6\times 101^{-19}\,{\rm C}$. Thus, the force is toward the negative $x$-axis.

Problem (5): The electric field due to charges $q_1=2\,\rm {\mu C}$ and $q_2=32\,\rm {\mu C}$ at distance $16\,\rm {cm}$ from charge $q_2$ is zero. What is the distance between the two charges?

Solution: Since the two charges $q_1$ and $q_2$ are positive, somewhere between them the net electric force must be zero that is at that point, the magnitude of the fields are equal (remember that the electric field of a positive charge at the field point is outward). Therefore, we get $\vec E_{net}=0 \quad \Longrightarrow \quad |\vec E_1|=|\vec E_2|$ Now use the definition of electric field to compute the elctric fields at that point due to the two other charges as below: $k\frac{|q_1|}{r_1^2}=k\frac{|q_2|}{r^{2}} \quad \Longrightarrow \quad \frac{2}{x^{2}}=\frac{32}{16^{2}}$ Taking square root of the both sides, we obtain $\frac{1}{x}=\frac{4}{16} \quad \Longrightarrow \quad x=4 \rm {cm}$
As shown in the figure, the distance of the two charges is $d=x+16=4+16=20\, \rm {cm}$.

Problem (6): In the figure, three  equal charges $q_1=q_2=q_3=+4\, \rm {\mu C}$ are located on the perimeter of a sphere of diameter $12\, \rm {cm}$. Find the net electric field, in terms of unit vectors $\hat i,\hat j$ at the center of the sphere. Solution:

Reasoning: The electric fields of charges $q_1$ and $q_3$ at the center of the sphere are equal in magnitude and opposite in direction. Since the magnitude of charges is the same $q_1=q_3$ and are located at an equal distance from the center so using the definition of the electric field we have $E_1=k\frac{|q_1|}{r^2}=E_3 \quad, \quad \vec E_1=-\vec E_3$ Therefore, the resultant of electric field vectors at point $\rm O$ is, using the superposition principle of fields, equal to the field $\vec E_2$.
$\vec E_{net,O}=\underbrace{\vec E_1+\vec E_3}_{0}+\vec E_2=\vec E_2$
But the electric field $\vec E_2$ lies in the fourth quadrant along the radius of the sphere which makes the angle of $53^\circ$ relative to the  $+x$ axis as shown in the figure. As mentioned already, in such cases we must decompose the vector into its components in $x$ and $y$ directions. First use the definition of electric field of a point charge to find the magnitude of $\vec E_2$ as \begin{align*} \vec E_2&=k\frac{|q_2|}{r^{2}}\\ &=\left(9\times 10^{9}\, \rm {N.\frac{m^2}{C^2}}\right)\frac{4\times 10^{-6}\, \rm C}{\left(\frac{12}{2}\times 10^{-2}\,\rm{m}\right)^2}\\ &=10^{7}\, \rm{\frac{N}{C}} \end{align*} From elementary geometry we can decompose the vector $\vec E_2$ as follows \begin{align*} \vec E_2&=\underbrace{|\vec E_2|\,\cos 53^\circ}_{E_{2x}}\left(\hat i\right)+\underbrace{|\vec E_2|\,\sin 53^\circ}_{E_{2y}}\left(-\hat j\right)\\ \\ &=10^{7}(0.6)\left(\hat i\right)+10^{7}(0.8)\left(-\hat j\right) \end{align*} By factoring and rearranging the above relation, we get
\begin{align*} \vec E_2&=10^{7}\left(0.6 \hat i-0.8 \hat j\right)\\ &=6 \times 10^{6} \hat i-8\times 10^{6} \hat j \quad \left({\rm \frac NC}\right)\end{align*}

Problem (7): As shown in the figure, the two charges $q_1$ and $q_2$ are fixed at the corners of the lower side of an isosceles triangle. If the electric field vector at point $A$ (in SI) is $\vec E_A=\left(7.2 \times 10^{4}\right)\hat i$, determine the type and magnitude of electric charges $q_1$ and $q_2$. Solution: The solution is straightforward. Use the superposition principle and find two relations between the magnitude of charges. Therefore, we have
\begin{align*}\vec E_A&=\vec E_1+\vec E_2 \\ \vec E_A&=k \frac{|q_1|}{r_1^2}\hat r_1+k \frac{|q_2|}{r_2^2}\hat r_2 \end{align*} Where in the second equality we have used the formula of the electric field of a point particle.

$\hat{r}$s are the unit vectors in an arbitrary direction (since we have no knowledge about being positive or negative of charges) that must be found as we proceed.

By decomposing unit vectors in $x$ and $y$ directions, and noting that in an isosceles triangle $r_1=r_2=d$, we have \begin{align*} \vec E_A&=\frac{k}{d^{2}}{|q_1|\left(\cos \alpha\, \hat i+\sin \alpha\, \hat j\right)+|q_2|\left(\cos \alpha \left(-\hat i\right)+\sin \alpha\, \hat j\right)}\\ \\ &=\frac{k}{d^2}{\cos \alpha\, \left(|q_1|-|q_2|\right)\hat i+\sin \alpha\, \left(|q_1|+|q_2|\right)\hat j}\end{align*} To decompose the unit vectors we have assumed the charges are positive.

Because the electric field at point $A$ is in the positive $x$ direction, so the $j$ component of the right hand side of the above, must be vanishes and its $i$ components must be equal to the left part as \begin{gather*} \vec E_A=\frac{k}{d^2}\,{\cos \alpha \left(|q_1|-|q_2|\right)\hat i+\underbrace{\sin \alpha \left(|q_1|+|q_2|\right)}_{0}\hat j}\\ \\ |q_1|+|q_2|=0\\ \\ 7.2\times 10^{4}=\frac{k}{d^2}\, \cos \alpha\, \left(|q_1|-|q_2|\right) \end{gather*} The first expression says that the magnitude of charges is opposite each other i.e. $|q_1|=-|q_2|$. From the second equation one can find the \begin{gather*} 2|q_1|\frac{k}{d^2}\,\cos \alpha =7.2\times 10^{4}\\ \Longrightarrow |q_1|=3.6\times 10^{4} \frac{d^2}{k\,\cos \alpha} \end{gather*} Use the Pythagorean theorem in the left triangle to find the value of $\cos \alpha$. \begin{gather*} 10^2=6^2+x^2 \quad \Longrightarrow \quad x=8\, \rm {cm}\\ \cos \alpha =\frac{8}{10} \end{gather*} By substituting the known data, we obtain \begin{align*}|q_1|&=3.6\times 10^4 \frac{\left(10\times 10^{-2} \rm {m}\right)^{2}}{\left(9\times 10^{9}\right)\left(0.8\right)}\\ \\ &=0.5\times 10^{-9}\, \rm {C}\\ \\ &=0.5\quad {\rm {nC}} \end{align*}Thus, $|q_1|=-|q_2|=0.5\, \rm {nC}$. Our initial assumption that the charges are positive does not correct. Therefore, we must choose correctly one of them to be positive and the other negative. By choosing $q_1$ to be positive and $q_2$ negative, one can arrive at the right net electric field at point $A$.

Problem (8): In the figure below, a point particle with a mass of $20\,{\rm g}$ and charge of $-3\,{\rm \mu C}$ is placed into a uniform electric field produced by two charged parallel plate. If we release the particle from the lower plate, at what speed the particle reaches the upper plate? Solution: Recall that a negatively charged particle moves in the opposite direction of the electric field lines. Two forces apply to the particle, one is electrostatic force, and the other is weight force.

The magnitude of the electric force acted on it is \begin{align*}F&=|q|E\\&=\left(3\times 10^{-6}\right)\left(2\times 10^{5}\right)\\&=0.6\quad {\rm N}\end{align*} and its weight is also $mg=\left(20\times 10^{-3}\right)(10)=0.2\,{\rm N}$ We can see the electric force is greater than the weight so the particle starts moving upwards. Next, we can use kinetic-energy theorem,$\Delta K=W_t$, to find the particle's speed at the upper plate. $W_t$ is the total work done on the particle and $\Delta K$ is the difference between initial and final kinetic energies. Thus, we have \begin{align*} W_t &=\Delta K \\ \\ W_F+W_{mg} &=\frac 12 mv_f^2-\frac 12 mv_i^2 \\ \\ Fd\,\cos \alpha + mgd\,\cos \theta&=\frac{1}{2}mv_f^2 +0 \\ \\ (0.6)(0.1)(1)+(0.2)(0.1)(-1)&=\frac 12 (20\times 10^{-3})v_f^2 \\ \\ 6\times 10^{-2}-2\times 10^{-2}&=10^{-2}v_f^2\\ \\ \Rightarrow v_f^2 &=4 \\ \\ v_f&=2\quad {\rm m/s}\end{align*} In above, we used the definition of work as $W=Fx \cos \theta$ where $x$ is the displacement vector and $\theta$ is the angle between the force and displacement.

In the third equality, $\alpha$ and $\theta$ are the angles between electric $F$ and weight $mg$ forces with displacement $d$, respectively.

For solving more problems on work in physics, refer to:
Physics work problems with solutions

Author: Ali Nemati
Last Update: 1/23/2021