Coulomb's Law: Solved Problems for High School and College
Practice problems with detailed solutions about Coulomb's law are presented that are suitable for the AP Physics C exam and college students. 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.
Coulomb's Law Practice Problems
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 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 the known numerical 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^{-16}\\ \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 signs 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 charge $q'=1\,{\rm \mu C}$.
(a) Find the magnitude of the Coulomb force that 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 (motionless) 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 the charges have opposite signs so the electric force between them is attractive.
if you are getting ready for AP physics exams, these electric force problems are also relevant.
Problem (3): What is the magnitude of the force that a ${\rm 25\, \mu C}$-charge exerts on a ${-\rm 10\,\mu C}$ charge ${\rm 8.5\, cm}$ away? (Take $k=9\times 10^{9}\,{\rm \frac{N.m^{2}}{C^2}}$)
Solution: Applying Coulomb's law, we can find the magnitude of the electric force as below \begin{align*}F&=k\frac{|qq'|}{r^2}\\&=(9\times 10^{9}){\rm \frac{(25\times 10^{-6}\,C)(10\times 10^{-6}\,C)}{(8.5\times 10^{-2}\,m)^2}}\\&=311.5\quad {\rm N}\end{align*} These two point charges have opposite signs, so the electrostatic force between them is attractive.
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Problem (4): Two charged particles apply an electric force of $5.2\times 10^{-3}\,{\rm N}$ on each other. Distance between them gets twice as much as before. What will be Coulomb's force?
Solution: Using Coulomb's law, we have $F=k\frac{|qq'|}{r^2}$, where $r$ is the distance between two charges. We are told in the problem that the distance is doubled so $r_2=2r_1$, thus the electric force is found as \begin{align*}F_2&=k\frac{|qq'|}{r_2^2}\\\\&=k\frac{|qq'|}{(2r_1)^2}\\\\&=\frac 14 \underbrace{k\frac{|qq'|}{r_1^2}}_{F_1}\\\\&=\frac 14 F_1\end{align*}
Problem (5): Two charged point particles are $4.41\,{\rm cm}$ apart. They are moved and placed in a new position. The force between them is found to have tripled. How far apart are they now?
Solution: initial distance is $r_1=4.41\,{\rm cm}$. At the new location, the force is tripled $F_2=3F_1$. Applying Coulomb's law, we have \begin{align*}F_2&=3F_1\\ \\ k\frac{\cancel{|qq'|}}{r_2^2}&=3k\frac{\cancel{|qq'|}}{r_1^2}\\ \\ \frac{1}{r_2^2}&=\frac{3}{(4.41\times 10^{-2})^2}\\\\ \Rightarrow r_2^2&=\frac{(4.41\times 10^{-2})^2}{3}\end{align*} Taking the square root of both sides, we get \[r_2=0.0254\,{\rm m}\] Thus, if those two charges are $2.54\,{\rm cm}$ away, the electrostatic force between them gets tripled.
Solve these practice problems on the electric charge to will get a better view of charges in physics.
The following problems are for practicing for the AP Physics C problems.
Problem (6): Three point charges are placed at the corners of an equilateral triangle as in the figure below. What is the magnitude and direction of the net electric force on the $2\,\rm \mu C$ charge?
Solution: The force that the charge $-6\,\rm \mu C$ applies to the $2\,\rm \mu C$ is attractive and to the right along the line connecting them and its magnitude is also calculated as below \begin{align*} F_{2,-6}&=k\frac{qq'}{r^2} \\\\ &=(9\times 10^9) \frac{(2\times 10^{-6})(6\times 10^{-6})}{(0.10)^2} \\\\ &=10.8\,\rm N \end{align*} The charge $8\,\rm \mu C$ is repelled the charge $2\,\rm \mu C$ along the line joining them with a magnitude of \begin{align*} F_{2,8}&=k\frac{qq'}{r^2} \\\\ &=(9\times 10^9) \frac{(2\times 10^{-6})(8\times 10^{-6})}{(0.10)^2} \\\\ &=14.4\,\rm N \end{align*} Given the geometry shown below, the force $\vec{F}_{2,8}$, denoted by $\vec{F}_8$ for simplicity, makes an angle of $60^\circ$ with the negative direction of $x$-axis.
Resolving this vector force along the horizontal and vertical directions gives its components \begin{align*} F_{1x}&=F_1 \cos 60^\circ \\ &=14.4\times (0.5)=72.2\,\rm N \\\\ F_{1y}&=F_1 \sin 60^\circ \\&=14.4\times (\frac{\sqrt{3}}{2})=72.2\sqrt{3}\,\rm N \end{align*} The net electric force on the charge $2\,\rm \mu C$ is the vector sum of individual forces due to other charges (superposition principle) \[\vec{F}_2=\vec{F}_8+\vec{F}_6 \] Adding the vectors along the horizontal and vertical directions give the corresponding components of the net electric force. \begin{align*} F_{2x}&=10.8+(-72.2) \\&=28.6\,\rm N \\\\ F_{2y}&=0+(-72.2\sqrt{3}) \\&=-72.2\sqrt{3} \,\rm N \end{align*} Given these components, we can find the magnitude and direction of the net electric force the desired charge \begin{align*} F_2&=\sqrt{F_{2x}^2+F_{2y}^2} \\\\ &=\sqrt{(28.6)^2+(-72.2\sqrt{3})^2} \\\\&=32.16\,\rm N \end{align*} and its direction with the positive $x$-direction as below \begin{align*} \alpha &=\tan^{-1}\left(\frac{F_{2y}}{F_{2x}}\right) \\\\ &=\tan^{-1}\left(\frac{72.2\sqrt{3}}{28.6}\right) \\\\ &=77^\circ \end{align*}
Problem (7): A $2-\rm \mu C$ point charge and another point charge of magnitude $4-\rm \mu C$ are a distance $L=1\,\rm m$ apart. Where should a third point charge be placed so that the net electric force on it is zero?
Solution: This is a tricky question and is more similar to the AP Physics C questions. We solve this problem by assuming the third point charge is positive.
Simple reasoning shows that between the charges the net force on the third charge points to the right and adds together rather than canceling each other. Thus, this region is removed.
Now, assume a point somewhere outside the charges and closer to the smaller one at distance $x$ from it. The following figure shows that the forces on this hypothetical positive charge are clearly in opposite directions.
\begin{gather*} F_2=F_4 \\\\ k\frac{q_3 q_2}{x^2}=k\frac{q_3 q_4}{(L+x)^2} \\\\ \frac{2}{x^2}=\frac{4}{(1+x)^2} \\\\ (1+x)^2=2x^2 \\\\ \Rightarrow \boxed{x^2-2x-1=0} \end{gather*} The above quadratic equation has two solutions \[x_1=2.41\,\rm m \quad , \quad x_2=-0.4\,\rm m\] The negative in the second solution means that we must go back to the region between the charges that is not acceptable. Thus, outside the charges and somewhere close to the smaller charge we can find a point where the net Coulomb force on the third charge is zero.
For practice, consider yourself a negative charge and repeat the above steps and find the result.
Problem (8): In the following figure, a small sphere of mass $3\,\rm g$ and charge of $q_1=15\,\rm nC$ are suspended by a light string over the second charge of equal mass and charge of $-85\,\rm nC$. The distance between the two charges is $2\,\rm cm$.
(a) Find the tension in the string?
(b) What is the smallest value of separation $d$, assuming the string can withstand a maximum tension of $0.150\,\rm N$?
Solution: The negative charge and gravity pull vertically down the positive charge and the tension in the string pulls up, as depicted in the following free-body diagram.
(a) The positive charge is motionless so the net force on it must be zero. Balancing the above forces applied to it gives the tension force in the string. \begin{gather*} T=F_e+mg \\\\ T=k\frac{q_1 q_2}{r^2}+mg \\\\ T=(9\times 10^9) \times \frac{(15\times 10^{-9})(85\times 10^{-9})}{(0.02)^2}+(0.003)(10) \\\\ \Rightarrow \boxed{T=0.31\,\rm N} \end{gather*}
(b) This time, the maximum tension force that the string can withstand is given and asked to find the smallest distance between the charges.
Again we use the above relation between the forces on the positive force and solve for the unknown distance $d$. \begin{gather*} T=k\frac{q_1 q_2}{r^2}+mg \\\\ T-mg=k\frac{q_1 q_2}{r^2} \\\\ r^2=k\frac{q_1 q_2}{T-mg} \end{gather*} Taking the square root of both sides and substituting the numerical values gives \begin{align*} r&=\sqrt{\frac{(9\times 10^9)(15\times 10^{-9})(85\times 10^{-9})}{0.150-(0.003)(10)}} \\\\ &=0.97\,\rm m \end{align*}
Problem (9): Four point charges are located on the corners of a square shown in the figure. If the net Coulomb force on $q_2$ is zero, what is the ratio of $\frac Qq$?
Solution: Since $|q_1|=|q_3|=q$ and placed at a equal distance of 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 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 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*}
On the following page, you find the properties of vectors:
Vector, definitions, formula, and solved-problems
Problem (10): In The configuration of three point 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 an 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 charge is given. See the later problem.
Problem (11): 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), 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 (12): 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 $B$ in 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 acting 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 (13): Three equal point charges are placed at the vertices of an equilateral triangle of side $a$. What is the magnitude and direction of the Coulomb force on the charge $q$ at the point $A$? ($q=10\,{\rm \mu C}$ and $a=\sqrt[4]{3}\,\rm m$).
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*} The $x$-components will add up to zero which gives the $x$-component of the net force on the charge on the position $A$. The 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 (14): Four unknown point 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 the superposition principle at point $4$ we get
\begin{gather*} \vec{F}_{net-on-q_4}=0 \\ \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 negative of the charges $q_1$ and $q_3$, 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). This result tells us that the force between $q_2$ and $q_4$ must be repulsive, or they must have like charges.
Because we assumed $q_4>0$, so $q_2$ is also positive. Thus, $\frac{q_2}{q_4}>0$ for this situation.
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 $\frac{q_2}{q_4}<0$.
Consequently, this analysis tells us that $q_2$ must be always positive.
Since $|q_1|=|q_3|=|q|$ and are at an equal distance to $q_4$ so their forces on $q_4$ due to these charges are also equal with magnitude (using Coulomb's law formula) \begin{align*} F_{14}=F_{34} &= \\ &= k\frac{|q_1 \, or \, q_3| |q_4|}{a^2}\\ &=k\,\frac{|q|\,|q_4|}{a^2} \end{align*} Pythagorean theorem gives the net electric force on $q_4$ due to $q_1$ and $q_3$ as \begin{align*} F&=\sqrt{F_{14}^2+F_{34}^2} \\\\ &=\sqrt{F_{14}^2+F_{14}^2} \\\\ &=\sqrt{2}\,F_{14} \end{align*} Now we proceed to determine the magnitude of $q_2$ by applying the equilibrium condition on charge $q_4$ (the magnitude of the forces along the square diagonal ($F$ and $F_{24}$) must be equal) 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{\left(5\times 10^{-6}\right)}{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. Therefore, the charge $q_2$ has a magnitude of $\boxed{10\sqrt{2}\,\rm \mu C}$.
Problem (15): 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 the 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 the $q_3$ cab be balanced and canceled by each other. Therefore, 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, the square root is taken from both sides. Solving the last equation for $x$, we get $x=10\,{\rm cm}$.
Problem (16): 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 $q_0=+20\,{\rm \mu C}$ and other charges be \begin{gather*} q_1=q_2=q_3=q_4=q_5=q_7=q_8=q=50\,{\rm \mu C} \\ q_6=-q \end{gather*} 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 $q_6$ attracts and $q_1$ repels 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 \[\vec{F}_O=18\,\left(\cos 45^\circ \, (-\hat{i})+\sin 45^\circ\,(-\hat{j})\right)\]
The following Coulomb's problems are for Honor Physics courses.
Coulomb's law Simple Solved Questions
Practice Problem (1): Suppose that two point charges, each with a charge of +1 Coulomb are separated by a distance of
1 meter.
(a) Will they attract or repel?
(b) Determine the magnitude of the electrical force between them.
Solution:
(a) Since the charges are like so the electric force between them is repulsive.
(b) The magnitude of electric force between two charges is found by Coulomb's law as below \begin{align*} F&=k\frac{|q_1 q_2|}{r^2}\\&=\big(9\times 10^9\big)\frac{1\times 1}{1^2}\\&=9\times 10^9\quad {\rm N}\end{align*}Where $|\cdots|$ denotes the absolute values of charges regardless of their signs.
Practice Problem (2): Two balloons are charged with an identical quantity and type of charge: -0.0025 C. They are held apart at a separation distance of 8 m. Determine the magnitude of the electrical force of repulsion between them.
Solution: applying Coulomb's law and putting the given numerical values in it, we have \begin{align*} F&=k\frac{q_1 q_2}{r^2}\\&=\big(9\times 10^9\big)\frac{(0.0025)(0.0025)}{(8)^2}\\&=879\quad {\rm N}\end{align*}
Practice Problem (3): Two charged boxes are 4 meters apart from each other. The blue box has a charge of +0.000337 C and is attracting the red box with a force of 626 Newtons. Determine the charge of the red box. Remember to indicate if it is positive or negative.
Solution: known information are $q_1=+0.000337\,{\rm C}$, $F=626\,{\rm N}$, and $d=4\,{\rm m}$. Unknown is $q_2=?$. Coulomb's law gets the magnitude of the force between two charges. By applying it and solving for $q_2$, we have \begin{align*} F&=k\frac{q_1 q_2}{d^2}\\ \\ \Rightarrow q_2&=\frac{F\,d^2}{k\,q_1}\\ \\ &=\frac{626\times (4)^2}{(9\times 10^9)(0.000337)}\\ \\&=0.0033\quad {\rm C}\end{align*} Since in the problem said that the force is attraction, so the charge of the red box must be negative.
Note that, Coulomb's law gives only the magnitude of the electric force without their signs.
Practice Problem (4): A piece of Styrofoam has a charge of -0.004 C and is placed 3.0 m from a piece of salt with a charge of -0.003 C. How much electrostatic force is produced?
Solution: the magnitude of the electrostatic force is determined as follows, \begin{align*} F&=k\frac{q_1 q_2}{d^2}\\&=\big(9\times 10^9\big)\frac{(0.004)(0.003)}{(3)^2}\\&=12000\quad {\rm N}\end{align*} Note that in Coulomb's force equation, the magnitude of the charges (regardless of their signs) must be included.
Practice Problem (5): Two coins lie 1.5 meters apart on a table. They carry identical electric charges. Approximately how large is the charge on each coin if each coin experiences a force of 2.0 N?
Solution: Substituting the given numerical into Coulomb's law equation, we have \begin{align*} F&=k\frac{q_1 q_2}{d^2}\\ \\ 2&=\big(9\times 10^9 \big)\frac{q\,q}{(1.5)^2}\\ \\ \Rightarrow q&=\sqrt{\frac{2\times (1.5)^2}{9\times 10^9}}\\ \\ &=2.23\times 10^{-5} \quad {\rm C} \end{align*}
In summary, Coulomb's law is basic for solving an electrostatic problem in the case of some arrangements of point charges. For offline reading, you can download this pdf version.
Author: Dr. Ali Nemati
Last Modified: Nov 28, 2022
