Practice problems with detailed solutions about Coulomb's law and electric force are presented which are suitable for high school and college students. These problems update regularly but for more solved problems (over 61) see here.

**Note:** In textbooks, the words Coulomb force, electric or electrostatic forces are interchangeably used for the force between two point charges.

**Problem (1):** Two like and equal charges are at a distance of $d=5\,{\rm cm}$ and exert a force of $F=9\times 10^{-3}\,{\rm N}$ on each other.

(a) Find the magnitude of each charge?

(b) What is the direction of the electrostatic force between them?

**Solution:** The magnitude of the force between two rest point charges $q$ and $q'$ separated by a distance $d$ is given by the Coulomb's law as below \[F=k\,\frac{|q|\,|q'|}{d^2}\] where $k \approx 8.99\times 10^{9}\,{\rm \frac{N.m^{2}}{C^2}}$ is the Coulomb constant and the magnitudes of charges denoted by $|\cdots|$.

Let the magnitude of charges be $|q_1|=|q_2|=|q|$, Now by substituting known values of $F$ and distance $d$ and solving for $|q|$ we get

\begin{align*} F&=k\,\frac{|q_1|\,|q_2|}{d^{2}}\\ 9\times 10^{-3}&=(8.99\times 10^{9})\frac{|q|^{2}}{(0.05)^{2}}\\ \Rightarrow q^{2}&=25\times 10^{-6}\\ \Rightarrow q&=5\times 10^{-8}\,{\rm C} \end{align*} In the second equality, we converted the distance from $cm$ to $m$ to coincide with SI units.

The direction of the Coulomb force depends on the sign of the charges. Two like charges repel and two unlike ones attract each other. Since $q_1$ and $q_2$ have the same sign so the electric force between them is repulsive.

**Problem (2):** A point charge of $q=4\,{\rm \mu C}$ is $3\,{\rm cm}$ apart from the charge $q'=1\,{\rm \mu C}$.

(a) Find the magnitude of the Coulomb force which one particle exerts on the other.

(b) Is the force attractive or repulsive?

**Solution:** Known values:

\begin{gather*}

|q|=4\,{\rm \mu C}\\ |q^{'}|=1\,{\rm \mu C}\\ d=3\,{\rm cm}=3\times 10^{-2}\,{\rm m}

\end{gather*}

(a) Coulomb's law gives the magnitude of the electric force between two stationary point charges so by applying it we have

\begin{align*}F&=k\,\frac{|q|\,|q'|}{d^{2}}\\&=(8.99\times 10^{9})\,\frac{(4\times 10^{-6})(1\times 10^{-6})}{(0.03)^2}\\&=40\,{\rm N}\end{align*}

(b) Since charges have opposite signs so the electric force between them is attractive.

**Problem (3):** Four point charges are located on the corners of a square as shown in the figure. If the net Coulomb force on $q_2$ be zero, then what is the ratio of $\frac Qq$?

**Solution:** Since $|q_1|=|q_3|=q$ and placed at a equal distance of the charge $q_2$ so $F_{12}=F_{32}$. We know that the resultant vector of two perpendicular and equal vectors $F$ is given as $\sqrt{2}\,F$ so, in this case, the magnitude of the net force acting on the charge $q_2$ due to $q_1$ and $q_3$ is $F=\sqrt{2}\,F_{12}$ along the diagonal ($q_2-q_4$) of the square and directed outward as shown in the figure.

The total electric force on the charge $q_2$ is the vector sum (superposition principle) of $\vec{F}_2=\vec{F}+\vec{F}_{42}$ since said that it is zero $\vec{F}_2=0$ so the electrostatic force of $q_4$ on $q_2$ i.e. $\vec{F}_{42}$ must be equal in magnitude and opposite in direction with $\vec{F}$. Therefore, by equating the magnitudes of the forces i.e. $F=F_{42}$ we obtain

\begin{align*}

F&=F_{42}\\ \sqrt{2}\,F_{12}&=F_{42}\\ \sqrt{2}\,k\,\frac{|q_1|\,|q_2|}{a^2}&=k\,\frac{|q_4|\,|q_2|}{(\sqrt{2}\,a)^2}\\ \sqrt{2}\frac{|q|\,|Q|}{1}&=\frac{|\frac 12 Q|\,|Q|}{2}\\\Rightarrow \frac Qq&=4\sqrt{2}

\end{align*}

**Problem (4):** Four point charges lie on the corners of a square of side $L=a\sqrt{2}$. What is the magnitude of the net Coulomb force at the place of charge $-q$?

**Solution:** Similar to the previous problem, first find (using the definition of Coulomb's law) the each electric force on the charge $-q$ then vector sum them and determine its magnitude. The charges $q_1$ and $q_3$ have the same magnitude and are at equal distances from $q_4$ so the magnitude of their Coulomb forces acted on $q_4$ are equal.\begin{align*}

\text{Coulomb's law}:\ F&=k\,\frac{|q|\,|q'|}{d^2}\\

F_{14}=F_{34}&=k\,\frac {|q|\,|-q|}{a\sqrt{2}}\\

&=k\,\frac{|q|^2}{2a^2}

\end{align*} Since the two charges have the opposite signs so the electric force (attractive) on $q_4$ due to $q_1$ is to the left and due to $q_3$ is downward as shown in the figure. Therefore, the net force of them, $F$, is $\sqrt{2}\,F_{14}$ or $\sqrt{2}\,F_{34}$.

Similarly, find the electrostatic force $F_{24}$ due to the charge $q_2$ on $q_4$.\begin{align*}F_{24}&=k\,\frac{|q|\,|-q|}{(2a)^2}\\

&=\frac 14\, \frac{k\,|q|^2}{a^2}\end{align*} The distance between $q_2$ and $q_4$ is the diagonal length of the square which is obtained using the Pythagorean theorem.

The charges $q_2$ and $q_4$ have opposite signs so the Coulomb force between them is attractive and directed inward along the diagonal of the square.

The magnitude of the net Coulomb force on $q_2$ is determined by adding the other magnitudes since they are directed in the same direction along the diagonal of the square. Thus,\begin{align*} F_2&=F+F_{24}\\ &=k\,\frac{|q|^2}{2a^2}+\frac 14\, \frac{k\,|q|^2}{a^2}\\ &=k\,\frac{|q|^2}{a^2}\,\left(\frac 12+\frac 14\right)\\ &= \frac 34 \, k\,\frac{|q|^2}{a^2}\end{align*}

**Problem (5):** In The configuration of charges, as shown in the figure below, the Coulomb force on each charge is zero. Determine the ratio of charges $q_3$ and $q_2$ i.e. $\frac {q_3}{q_2}$.

**Solution:** Since the ratio of the $\frac {q_3}{q_2}$ is required and the net force on each charges is zero so we must balance the forces on the charge $q_1$ because in this case the magnitude of $q_1$ cancels from both sides as below,\begin{align*}F_{21} &= F_{31} \\

k\,\frac{|q_1|\,|q_2|}{(20)^2} &=k\,\frac{|q_1|\,|q_3|}{(30)^2}\\ \frac{|q_2|}{400}&=\frac{|q_3|}{900}\\ \Rightarrow \frac{|q_3|}{|q_2|}&=\frac 94 \end{align*} Note: since the expression above is a equality so no need to convert the units to SI.

Now that the ratio of the magnitudes of the charges is obtained so we must determine its signs. As you can see in the figure, because the forces $\vec{F}_{31}$ and $\vec{F}_{32}$ are in the opposite directions (to produce a zero net force on $q_1$) so the charges $q_2$ and $q_3$ must be unlike.

The exact sign of charges can not be determined as long as at least the sign of one the charge is given. See the later problem.

**Problem (6):** Two point charges $q_1=+2\,{\rm \mu C}$ and $q_2=+8\,{\rm \mu C}$ are $30\,{\rm cm}$ apart from each other. Another charge $q$ is placed so that the three charges are brought to a balance. What is the magnitude and sign of the charge $q$?

**Solution:** To find the location of the third charge, place a positive (or negative) test charge $q_3$ somewhere between $q_1$ and $q_2$. Since all charges here are positive (negative) so, by Coulomb's law, the electrostatic forces on the test charge are repulsive (attractive) and to the left (right) and right (left) of it. Consequently, the net electric force can be zero between them at a distance of say $x$ from charge $q_1$.

Now, balance the magnitude of the forces on the test charge $q_3$ as below to find the location of it \begin{align*}F_{13}&=F_{23}\\ k\,\frac{|q_1|\,|q_3|}{x^2} &=k\,\frac{|q_2|\,|q_3|}{(30-x)^2}\\ \frac {2}{x^2}&=\frac {8}{(30-x)^2}\\ \Rightarrow 2x&=30-x\\ \Rightarrow x&=10\,{\rm cm} \end{align*} In above, the required charge $q_3$ is canceled from both sides and one can not find its sign and value. To find the magnitude and sign of $q_3$, balance the forces on another charge, say $q_1$ as below

\begin{align*} F_{31}&=F_{21}\\ k\,\frac{|q_1|\,|q_3|}{(10)^2} &=k\,\frac{|q_2|\,|q_1|}{(30)^2}\\ \frac {|q_3|}{100}&=\frac {8}{900}\\

\Rightarrow |q_3|&=\frac 89\\ \end{align*}

The electric force $\vec{F}_{21}$ is repulsive and directed to the $-x$ axis. Since the net force on each charge is zero so the charge $q_3$ must be negative to provide an attraction force in the opposite direction of $\vec{F}_{21}$ that is to the $+x$ axis.

Therefore, the third charge is negative, located at a distance of $10\,{\rm cm}$ between the two other charges.

**Problem (7):** In the corners of a square of side $L$, four point charges are fixed as shown in the figure below. What angle does make the net Coulomb force vector on the charge $q$ located at the point $B$ on the upper right corner with the horizontal?

**Solution:** In this problem, there is no need to do any explicit calculation, only justify the desired direction.

The electric force vector on the charge $q$ at the corner $B$ is the vector sum of the forces acted by the other charges $-q$ on it. Therefore, using superposition principle, we have \[\vec{F}_B=\vec{F}_{AB}+\vec{F}_{DB}+\vec{F}_{CB}\]

Similar to the previous problems, since the magnitude and distance of charges located at $A$ and $C$ are equal and the same so $|\vec{F}_{AB}|=|\vec{F}_{CB}|=F$. On the other hand, those forces are attractive and directed to the points $A$ and $C$ as shown in the figure. Thus, their resultant electric force lies along the diagonal of $BD$ points inward with the magnitude of $\sqrt{2}\,F$.

The force between charge $-q$ at point $D$ and $q$ at point $B$ is also attractive, lies along the diagonal of $BD$ and points inward. Therefore, Adding these three force vectors gives a resultant Coulomb force vector $\vec{F}_B$ directed with an angle of $(180+45)^\circ$ along the $BD$ diagonal as shown in the figure.

**Problem (8):** Three equal charges are placed at the vertices of an equilateral triangle of side $L$. What is the magnitude and direction of the Coulomb force on the charge $q$ at the point $A$? ($q=10\,{\rm \mu C}$ and $a=\sqrt[4]{3}$).

**Solution:** first find the magnitudes of $\vec{F}_{BA}$ and $\vec{F}_{CA}$ using Coulomb's force law as below

\begin{align*} F_{BA}&=k\,\frac{|q_B|\,|q_A|}{d^2}\\ &=k\,\frac{q^2}{\left(\sqrt[4]{3}\right)^2}\\ &=k\,\frac{q^2}{\sqrt{3}}\\

&=(9\times 10^{9})\,\frac{(10\times 10^{-6})^2}{\sqrt{3}}\\ &=\frac{9}{\sqrt{3}}\times 10^{-1}\,{\rm N} \end{align*} Since the distance to $q_A$ and the magnitudes of $q_B$ and $q_C$ are the same so $F_{BA}=F_{CA}=F$.

Now find the direction of the electrostatic forces above using vector components. $\vec{F}_{BA}$ makes an angle of $60^\circ$ with the $+x$ direction and $\vec{F}_{CA}$ an angle of$-60^\circ$ with the $-x$ direction. Thus, the above forces can be written in the following vector form

\begin{align*}

\vec{F}_{BA}&=\underbrace{|\vec{F}_{BA}|}_{F}\left(\cos 60^\circ\,\hat{i}+\sin 60^\circ \,\hat{j}\right)\\

\vec{F}_{CA}&=\underbrace{|\vec{F}_{CA}|}_{F}\left(\cos 60^\circ\, (-\hat{i})+\sin 60^\circ \,\hat{j}\right)

\end{align*}

Sum the $x$-components to find the resultant $F_{Ax}$ which gets zero. Sum of the $y$-components also gives

\begin{align*} F_{Ay} &= F\,\sin 60^\circ+F\,\sin 60^\circ \\ &=2F\,\sin 60^\circ \\ &=2F\times\left(\frac {\sqrt{3}}{2}\right)\\ &=\sqrt{3}\,F\\ &=\sqrt{3}\times\frac{9}{\sqrt{3}}\times 10^{-1}\\ &=0.9\,{\rm N} \end{align*}

Therefore, the resultant Coulomb force on $q_A$ directed upward and is written as $\vec{F}_A=0.9\,\hat{j}$.

**Problem (9):** Four unknown charges are held at the corners of a square. Suppose $q_4$ is at equilibrium and Let $q_1=q_3=-5\,{\rm \mu C}$ then what is the magnitude of the charge $q_2$ and the sign of the ratio of $\frac {q_2}{q_4}$.

**Solution:** Since $q_4$ is at equilibrium so the net electric force on it must be zero. Applying superposition principle at point $4$ we get

\begin{gather*} \vec{F}_4=\vec{F}_{14}+\vec{F}_{24}+\vec{F}_{34}=0\\ \Rightarrow \vec{F}_{14}+\vec{F}_{34}=-\vec{F}_{24} \end{gather*}

Let's consider first the charge $q_4$ is positive. Because of being positive of the charges $q_1$ and $q_3$, so their forces on $q_4$ are attractive, to the right and up direction which gives a net force $F$ along the diagonal of the square and directed inward. In this case, the electric force $\vec{F}_{24}$ must be diagonally and directed outward to cancel the contribution $F$ (See the right figure). Similar reasoning can be also applied for the case of a negative $q_4$ charge (left figure). Consequently, $q_4$ and $q_2$ are unlike charges or its ratio is $\frac{q_2}{q_4}<0$.

Since $|q_1|=|q_3|=|q|$ and are at the equal distance to $q_4$ so the their forces on $q_4$ due to these charges are also equal with magnitude (using Coulomb's law formula) \[F_{14}=F_{34}=k\,\frac{|q|\,|q_4|}{a^2}\]. Pythagorean theorem gives the net electric force on $q_4$ due to $q_1$ and $q_3$ as $F=\sqrt{2}\,F_{14}$.

Now we proceed to determine the magnitude of $q_2$ by applying the equilibrium condition on the charge $q_4$ as below

\begin{align*}

F&=F_{24}\\

\sqrt{2}\,F_{14}&=k\,\frac{|q_2|\,|q_4|}{(\sqrt{2}\,a)^2}\\

\sqrt{2}\,k\,\frac{|q_1|\,|q_4|}{a^2}&=k\,\frac{|q_2|\,|q_4|}{(\sqrt{2}\,a)^2}\\

\sqrt{2}\,\frac{5\times 10^{-6}}{a^2}&=\frac{|q_2|}{2a^2}\\

\Rightarrow |q_2|&=10\sqrt{2}\,{\rm \mu C}

\end{align*}

the first equality is the equilibrium condition.

**Problem (10):** Two point charges of $q_1=+2\,{\rm \mu C}$ and $q_2=-8\,{\rm \mu C}$ are at a distance of $d=10\,{\rm cm}$. Where must a third charge $q_3$ be placed so that the net Coulomb force acted upon it is zero?

**Solution:** Put a positive (or negative) test charge $q_3$ between them and examine whether net Coulomb force on it is zero or not. In this case, the net electrostatic force on the positive (negative) test charge due to the charges $q_1$ and $q_2$ is to the right (left). Thus, there is no space between them to balance a test charge.

Now place that test charge $q_3$ outside them, say in the left of the charge $q_1$ at distance $x$ from it. One can see that, in this case, the forces on $q_3$ balance and can be canceled each other. Therefor, apply Coulomb's force law and find the unknown $x$ as below,

\begin{align*}

F_{13}&=F_{23}\\

k\,\frac{|q_1|\,|q_3|}{x^2} &=k\,\frac{|q_2|\,|q_3|}{(10+x)^2}\\

\frac {|2\times 10^{-6}|}{x^2}&=\frac{|-8\times 10^{-6}|}{(10+x)^2}\\

\frac {1}{x^2}&=\frac{4}{(10+x)^2}\\

\Rightarrow \frac 14 &=\frac {x^2}{(x+10)^2}\\

\Rightarrow \frac{x}{x+10}&=\pm \frac 12

\end{align*}

In the fifth equality, square root is taken from both sides. Solving the last equation for $x$, we get $x=10\,{\rm cm}$.

**Problem (11):** In the figure below, what is the magnitude and direction of the net Coulomb force vector acted on the charge $q_O=q$ by the eight other charges placed on the circumference of a circle of radius $R=100\,{\rm cm}$.

Let $q_0=+20\,{\rm \mu C}$ , $q_1=q_2=q_3=q_4=q_5=q_7=q_8=q=50\,{\rm \mu C}$ and $q_6=-q$. The charge $q_0$ is held at the center of circle.

**Solution:** Using the symmetry of the charge configuration, one can realize that the electric forces due to pair of charges $(q_1,q_5)$ , $(q_2,q_8)$ and $(q_3,q_7)$ on the charge at the origin $q_O$ are equal in magnitude and opposite in direction, so cancel each other. Consequently, the net force on the charge $q$ at the center is only due to the charges $q_6$ and $q_2$ which its magnitudes ($F_{1O}$ and $F_{6O}$) are computed by applying Coulomb's law as below

\begin{align*} F&=F_{1O}=F_{6O}\\ F&=k\,\frac{|q_1|\,|q|}{R^2}=k\,\frac{|q_6|\,|q|}{R^2}\\ F&=k\,\frac{|q|\,|q|}{R^2}=k\,\frac{|-q|\,|q|}{R^2}\\

\Rightarrow F&=k\,\frac{|q|^2}{R^2}\\ &=(9\times 10^{9})\, \frac{(50\times 10^{-6})(20\times 10^{-6})}{(100\times 10^{-2})^2}\\ &=9\,{\rm N}

\end{align*}

The charge $q_6$ attract and $q_1$ repel the charge $q$ at the center so the magnitude of the net electric force at point $O$ is $2$ times the magnitude of the force between $q_6$ or $q_1$ and $q$ at center i.e. $|\vec{F}_O|=2F=19\,{\rm N}$.

The resultant electric force $\vec{F}_O$ lies on the third quadrant, points radially outward and makes an angle of $(180+45)^\circ$ with the positive $x$ axis or $45^\circ$ with the $-x$ axis. Its vector form is written as follows \[\vec{F}_O=18\,\left(\cos 45^\circ \, (-\hat{i})+\sin 45^\circ\,(-\hat{j})\right)\]

In summary, Coulomb's law is basic for solving an electrostatic problem in the case of some arrangements of point charges.

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