# Electric Potential Problems and Solutions for AP Physics 2

Several problems on electric potential are provided with detailed solutions. All these questions are for high schools and AP Physics exams.

## Electric Potential Problems:

Problem (1): How much work does the electric field do in displacing a proton from a position at a potential of $+120\,\rm V$ to a point at $-35\,\rm V$? Express your result both in joules and electron-volt.

Solution: the potential difference is defined as the work done per unit charge to move a point charge from one point to another $V_2-V_1=\frac{W}{q}$ The SI unit of potential is the volt ($\rm V$).

In this question, the electric potential at two different points are given and asked the amount of work done on the proton with a charge of $q=+1.6\times 10^{-19}\,\rm C$. \begin{align*} W&=q(V_2-V_1) \\ &=(1.6\times 10^{-19})\,(-35-120) \\&=\boxed{-2.48\times 10^{-17}\,\rm J} \end{align*} Electron-volt is another way to measure energy at the microscopic level. To convert the joules into the electronvolt, we use the following formula $1\,\rm eV=1.6\times 10^{-19}\,\rm J$ Thus, dividing the joules by the electron charge magnitude, we can obtain electronvolt unit. Consequently, $\frac{2.48\times 10^{-17}}{1.6\times 10^{-19}}= \boxed{155\,\rm eV}$

Problem (2): A proton is accelerated from rest through a potential difference of $120\,\rm V$.
(a) How fast does the proton move across this region?
(b) How fast does an electron move through the same potential difference?

Solution: This type of question appears in all the electric potential problems. To solve such problems, keep in mind that, you must first find the kinetic energy $K=\frac 12 mv^2$ of the accelerated point charge through the given potential difference. Then equate that with the work done on the point charge by the electric forces, $W=q\Delta V$. \begin{gather*} K=W=q\Delta V \\\\ \frac 12 mv^2=q\Delta V \\\\ \Rightarrow \boxed{v=\sqrt{\frac{2q\Delta V}{m}}}\end{gather*}
(a) Substitute the numerical values related to the mass and charge of the proton into the above formula \begin{align*} v&=\sqrt{\frac{2q\Delta V}{m}} \\\\&=\sqrt{\frac{2(1.6\times 10^{-19})(120)}{1.67\times 10^{-27}}} \\\\&= 152\times 10^{3}\,\rm m/s \end{align*}
(b) To find the speed of an electron accelerated from rest through a potential difference $\Delta V$, put the numerical values of an electron into the above expression \begin{align*} v&=\sqrt{\frac{2q\Delta V}{m}} \\\\&=\sqrt{\frac{2(1.6\times 10^{-19})(120)}{9.11\times 10^{-31}}} \\\\&= 6.42\times 10^{6}\,\rm m/s \end{align*}

Problem (3): Suppose an electron moving at a constant speed of $8.4\times 10^5\,\rm m/s$. What potential difference is needed to stop it?

Solution: here the initial and final kinetic energies of the electron are given. The magnitude of decrease in the kinetic energy is equal the work done by the electric forces or $\Delta K=\underbrace{q\Delta V}_{W}$ where $\Delta K=\frac 12 m(v_2-v_1)^2$. Substituting the numerical values of the electron into above and solving for $\Delta V$, we will get \begin{align*} \Delta V &=\frac{\frac 12 m(v_f^2-v_i^2)}{e} \\\\ &=\frac{(9.11\times 10^{-31})(0-(8.4\times 10^5)^2)}{2(-1.6\times 10^{-19})} \\\\ &=2 \quad\rm V \end{align*} where the electric charge of electron is $q=-e$.

Problem (4): The kinetic energy of a proton is $4.2\,\rm keV$. How fast does the proton move?

Solution: the kinetic energy is defined $K=\frac 12 mv^2$ and its SI unit is $\rm J$. But in this case, it is given in $\rm eV$. Thus, first multiplying it by the electron charge magnitude to convert it in joules \begin{align*} 4.2\,\rm keV&=4.2\times 10^3\times (1.6\times 10^{-19}) \\&=6.72\times 10^{-16}\,\rm J\end{align*} Now substitute everything into the kinetic energy formula and solve for $v$ \begin{align*} v&=\sqrt{\frac{2K}{m}} \\\\ &=\sqrt{\frac{2(6.72\times 10^{-16})}{1.67\times 10^{-27}}} \\\\ &=898\,\rm m/s \end{align*}

Problem (5): The potential difference between two parallel plates $7.5\,\rm mm$ apart is $240\,\rm V$. What electric field strength does form between them?

Solution: The magnitude of the electric potential difference between two points in a uniform electric field $E$ is found by $\Delta V=Ed$ where $d$ is the distance between the two points. We assume in a region away from the edges of the two parallel plates, the electric field is uniform. Thus, at that point the electric field magnitude is $E=\frac{\Delta V}{d}=\frac{240}{7.5\times 10^{-3}}=32000\,\rm V/m$

Problem (6): To move a charge of $-5\,\rm \mu C$ from point A to point B, a work of $9.5\times 10^{-4}\,\rm J$ is done by an unknown external force. The charge initially is at rest and finally acquires kinetic energy of $3.5\times 10^{-4}\,\rm J$ at point B.
(a) How large is the potential difference between A and B?
(b) Compare the potentials at points $A$ and $B$.

Solution: According to the work-kinetic energy theorem, the work done on an object between two points by some forces could cause that object to gain kinetic energy $\Delta K=W$ (a) In this problem, the work done by the external force is not equal the kinetic energy of the charge at point B. This indicates that there must be another work, such as work done by an electric force $W_E$, that has not been given. $\Delta K=W_F+W_E$  On the other hand, we know that the work done by an electric force equals $W_E=-q\Delta V$. Combining these two equations and solving for $\Delta V$, we get \begin{gather*} K_f-K_i=W_F-q\Delta V \\\\ \Rightarrow \Delta V= \frac{K_f-K_i-W_F}{-q} \\\\ = \frac{(3.5\times 10^{-4})-(9.5\times 10^{-4})}{-(-5\times 10^{-6})} \\\\ =\boxed{-120\quad \rm V} \end{gather*} where we set the initial kinetic energy $K_i=0$, since the charge is initially at rest, $v_i=0$.
(b) The potential difference between $A$ and $B$ is found $V_B-V_A=-120\,\rm V$ This tells us that point A is at a higher electric potential or $V_A>V_B$.

This question was similar to the work-kinetic energy problems.

Problem (7): A $+9\,\rm \mu C$ charge moves from the origin to a point of coordinate $(x=2\,\rm cm,y=5\,\rm cm)$ in a uniform electric field of magnitude $240\,\rm V/m$. Through what electric potential difference did the charge move?

Solution: The magnitude of the electric potential difference $\Delta V$ and the electric field strength $E$ is related together by the formula $\Delta V=Ed$ where $d$ is the distance between the initial and final points.

In this case, the initial point is located at origin $x_i=(0,0)$ and the final point is at $x_f=(2,5)$. Distance is the magnitude of displacement $\Delta x$. According to solved examples of displacement, we have $\Delta x=x_f-x_i=(2,5)$ Displacement is a vector quantity in physics whose magnitude is found using Pythagorean theorem $d=\sqrt{2^2+5^2}=5.4\, \rm m$ Thus, those two points are separated by $5.4\,\rm m$ in a uniform electric field, so the potential difference between these points is $\Delta V=Ed=240\times 5.4=1296\,\rm V$

Problem (8): A particle having charge $q=-3.6\,\rm \mu C$ and mass $m=0.045\,\rm kg$ has an initial velocity $v_i=20\,\rm m/s$ at the origin. At a point $2\,\rm m$ farther, its speed reduced to $v_f=10\,\rm m/s$. What is the potential difference between the origin and that point? Which point is at the lower potential?

Solution: We can see this as a work problem in physics which is done by some forces. According to the work-kinetic energy theorem, the work done on an object equals the change in the kinetic energy of that object, i.e., $W=\Delta K$.

On the other hand, the work required to displace a charged object of $q$ between two points with potential difference $\Delta V=V_2-V_1$ is $W=-q\Delta V$. By combining these two later statements, we arrive at the following conclusion $-q\Delta V=\Delta K$ we can see this as the work-kinetic energy theorem for electrostatic.

Substituting the numerical values into the above expression, we have \begin{gather*}-q(V_f-V_i)=\frac 12 m(v_f^2-v_i^2) \\\\ -(-3.6) \Delta V=\frac{0.045\times \left(10^2-20^2\right)}{2} \\\\ \Rightarrow \quad \Delta V=\frac{-6.75}{3.6}=-1.875\,\rm V \end{gather*} we can use this potential difference and determine which point is at higher or lower potential. By definition, the difference between the final and initial potentials, called potential difference $\Delta V$, is \begin{gather*} \Delta V=V_f-V_i \\ -1.875 =V_f-V_i \\ \Rightarrow \quad \boxed{V_f=V_i-1.875} \end{gather*} Therefore, the potential at a point $2\,\rm m$ away from the origin is lower than the potential at the origin. $V_f<V_i$

Problem (9): A particle with a charge of $+3.6\,\rm nC$ is released from rest in a uniform electric field $\vec{E}$ directed to the right. Its kinetic energy is measured to be $1.8\times 10^{-6}\,\rm J$ after traveling a distance of $4\,\rm cm$.
(a) What work was done on the particle by the electric field?
(b) Compare the potentials of the starting and end points.

Solution: The work done by the electric force on a charged object is calculated by the formula $W=-\Delta V$, where $\Delta V=V_f-V_i$ is the potential difference between those two points where the particle travels.

(a) In this part, we can simply use the work-kinetic energy theorem, $\Delta K=W$, and find the work done by the electric force. \begin{align*} W&=K_f-K_i \\ &=1.8\times 10^{-6} \quad \rm J \end{align*} where we set $K_i=0$ because $v_i=0$.

(b) These work done by the electric force equals $W=-q\Delta V$. Thus, the potential difference between initial and final points is \begin{align*} V_f-V_i&=\frac{W}{-q} \\\\ &= \frac{1.8\times 10^{-6}}{-3.6\times 10^{-9}} \\\\ &=-500\,\rm V \\\\ \Rightarrow \quad \boxed{V_f=V_i-500 }\end{align*}The above statement tells us that the end potential is $500\,\rm V$ less than the start potential, as expected. Because the object is positively charged and is released from rest in an electric field, it follows a path from high potential to low potential such as when you release an object at a gravitational potential.

## Potential Due to Point Charges: Problems

Problem (10): What is the electric potential at a distance of $2\,\rm cm$ from a proton?

Solution: The electric potential due to a point charge $q$ at distance $r$ from that is calculated by the formula $V=k\frac{q}{r}$ where $k=8.99\times 10^9 \,\rm \frac{N\cdot m^2}{C^2}$ is the coulomb's constant. The electric charge of proton is $q=+e=1.6\times 10^{-19}\,\rm C$. \begin{align*} V&=(8.99\times 10^9) \frac{1.6\times 10^{-19}}{0.02} \\\\ &=7.2\times 10^{-8}\,\rm V \end{align*}

Problem (11): What is the electric potential at a distance of $3\,\rm cm$ from a charge of $q=-1.5\,\rm nC$?

Solution: Substitute the numerical values into the electric potential formula due to point charge $q$ at distance $r$ from it. \begin{align*} V&=(8.99\times 10^9) \frac{-1.5\times 10^{-9}}{0.03} \\\\ &=-449.5\,\rm V \end{align*} Note that, unlike electric field problems or Coulomb's law problems where the sign of charges is not included in the calculation, the sign of charges must be included for the electric potential.

Problem (12): Suppose two identical point charges of $2\,\rm \mu C$ positioned at equal distance from a positive test charge $q=1.5\times 10^{-18}\,\rm C$ located at the origin, as shown in the figure below. Find the electric potential at the origin due to the two $2-\rm \mu C$ charges.

Solution: keep in mind that the electric potential is a scalar quantity as opposed to electric field and force. To find the potential at a point, first, find the potential due to each charge at the desired point, then simply add up all the previous contributions.

In this case, the two $2-\rm \mu C$ charges are at the equal distance of the point of interest (origin). Thus, their contributions in the potential at that point is \begin{align*} V&=k\frac{Q}{r}\\\\ &=(8.99\times 10^9)\frac{2\times 10^{-6}}{0.8} \\\\ &=22475\,\rm V \end{align*} Now, we simply sum these potentials due to those two charges to find the potential at origin. $V_{tot}=V_2+V_2=2\times 22475=44950\,\rm V$

Problem (13): Suppose two identical point charges of $25\,\rm \mu C$ are located $50\,\rm cm$ away from each other. How much work would be required to displace a test point charge of $0.2\,\rm \mu C$ from a point midway between them to a point $12\,\rm cm$ closer to either of the charges?

Solution: The work required to move a point charge in the presence of the electric potential of other charges is calculated by $W=-q\Delta V$ where $\Delta V=V_f-V_i$ is the potential difference created by other charges between the initial and final points.

In this case, the initial point is in the middle of two identical charges. The electric potential at this point is simply sum of the potentials due to each $25-\rm \mu C$ charge. $V_i=V_{25}+V_{25}$ where \begin{align*} V&=k\frac{Q}{r}\\\\ &=(8.99\times 10^9)\frac{25\times 10^{-6}}{0.25} \\\\ &=8.99\times 10^5\,\rm V \end{align*} Hence, the total electric potential at the initial point is $V_i=1.8\times 10^6\,\rm V$ It is said that the final location is $12\,\rm cm$ closer to either of charges. As a result, that point is placed $12\,\rm cm$ from ,say right charge, and $50-12=38\,\rm cm$ from the left charge as shown in the figure. The potential due the left charge at this point is \begin{align*} V_L&=k\frac{Q}{r}\\\\ &=(8.99\times 10^9)\frac{25\times 10^{-6}}{0.38} \\\\ &=0.6\times 10^6\,\rm V \end{align*} and for the right charge is \begin{align*} V_R&=k\frac{Q}{r}\\\\ &=(8.99\times 10^9)\frac{25\times 10^{-6}}{0.12} \\\\ &=1.9\times 10^6\,\rm V \end{align*} Therefore, the total electric potential at the final point is $V_f=V_L+V_R=2.5\times 10^6\,\rm V$ Now, we must calculate the change in the potential of these two locations as $\Delta V=V_f-V_i=0.7\times 10^6\,\rm V$ The given test charge wants to move in this potential difference, so the work required is calculated as follows \begin{align*} W&=-q\Delta V \\\\ &=-(0.2\times 10^{-6})(0.7\times 10^6) \\\\ &=-0.14\,\rm J \end{align*} The negative indicates that an external force must do work to move a positive test charge in this charge configuration.

Problem (14): In the figure below, two charges are placed at a distance of $d=6\,\rm cm$. What is the electric potential at (a) a point in the middle of the two charges? (b) point $A$ as shown. (Take $k=9\times 10^9\,\rm N\cdot m^2/C^2$)

Solution: In this problem, two charges are positioned at the corners of the base of an equilateral triangle. The electric potential at an arbitrary point at a distance $r$ from a point charge is $V=k\frac{q}{r}$.

(a) At point $B$, in the middle of the two charges, the electric potential is the sum of individual potentials due to each charge. $V_B=V_{15}+V_{-8}$ where \begin{align*} V_{15}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{15\times 10^{-9}}{0.03} \\\\ &=4.5\,\rm V \end{align*} and the potential due to $-8\,\rm nC$ is \begin{align*} V_{-8}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{-8\times 10^{-9}}{0.03} \\\\ &=-2.4\,\rm V \end{align*} Hence, the potential at point $B$ is $V_B=4.5+(-2.4)=2.1\,\rm V$ We emphasize again that the sign of electric charges must be included in the formula of electric potential.

(b) Similarly, the potential at point $A$ is $V_A=V_{15}+V_{-8}$ where \begin{align*} V_{15}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{15\times 10^{-9}}{0.06} \\\\ &=2.25\,\rm V \end{align*} and  \begin{align*} V_{-8}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{-8\times 10^{-9}}{0.06} \\\\ &=-1.2\,\rm V \end{align*} Therefore, we have $V_A=2.25+(-1.2)=1.05\,\rm V$

Problem (15): A charge of $+2q$ is at origin and another point charge of $-4q$ is on the position $x=10\,\rm cm$. At what point on the $x$-axis is the electric potential zero?

Solution: First, draw a coordinate axis and place the charges on the given coordinates as shown in the figure below. Now, Assume the potential is zero, for example, at an arbitrary point at a distance of $x$ from the origin and outside the charges, say $A$.

By calculating the net electric potential due to those charges at that imaginary point and setting it to zero, gives \begin{gather*} V_A =k\frac{2q}{x}+k\frac{-4q}{x-0.10} \\\\ 0=k\left(\frac{2q}{x}-\frac{4q}{x-0.10}\right) \end{gather*} We cancel out the $q$ on each side and solve for $x$ to find $\boxed{x=-0.1\quad\rm m}$ The minus sign indicates that the point $A$ must be chosen to the left of the origin at a distance of $10\,\rm cm$ (as shown in the figure below).

As a result, outside the charges and at a distance of $10\,\rm cm$ closer to the smaller charge, the electric potential is zero.

Problem (16): Find the electric potential at the left upper corner of the rectangle in the following figure.

Solution: Due to the scalar nature of the electric potential, all we should do is that find individual potentials due to each charge at that specific point, then simply add them up together. The potential due to the charge $4\,\rm \mu C$ at point $A$ is  \begin{align*} V_4&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{4\times 10^{-6}}{0.03} \\\\ &=1.2\times 10^6\,\rm V \end{align*} For the charge $1\,\rm \mu C$, we have  \begin{align*} V_1&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{1\times 10^{-6}}{0.02} \\\\ &=0.45\times 10^6\,\rm V \end{align*} And similarly, for the charge $-2\,\rm \mu C$,  \begin{align*} V_{-2}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{-2\times 10^{-6}}{0.05} \\\\ &=-0.36\times 10^6\,\rm V \end{align*} Note that to find the distance of the charge $-2\,\rm \mu C$ from the point $A$, we applied the Pythagorean theorem as below $\sqrt{2^2+3^2}=5\,\rm cm$ Next, we simply add all these potentials together. \begin{gather*} V_A=(1.2+0.45-0.36)\times 10^6 \,\rm V \\\\ \Rightarrow \boxed{V_A=1.29\times 10^6\,\rm V}\end{gather*}

Problem (17): In the following configuration, what is the electric potential difference $V_A-V_B$?

Solution: At point $A$ the potential due to the given charge is obtained using the the electric potential formula $V=k\frac{q}{r}$ \begin{align*} V_A&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{1\times 10^{-6}}{2} \\\\ &=4500\,\rm V \end{align*} and at point $B$, we have \begin{align*} V_B&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{1\times 10^{-6}}{1} \\\\ &=9000\,\rm V \end{align*} Thus, the potential difference $V_A-V_B$ is $V_A-V_B=4500-9000=-4500\,\rm V$

Problem (18): Point $B$ is at $1\,\rm m$ north of a $1\,\rm \mu C$ charge, and point $A$ is east of the charge at a distance of $2\,\rm m$ from it. What is the potential difference $V_A-V_B$?

Solution: Similar to the previous problem, potential at point $A$ is \begin{align*} V_A&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{1\times 10^{-6}}{2} \\\\ &=4500\,\rm V \end{align*} and the potential at point $B$ is \begin{align*} V_B&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{1\times 10^{-6}}{1} \\\\ &=9000\,\rm V \end{align*} Thus, we have $V_A-V_B=4500-9000=-4500\,\rm V$

Problem (19): Two point charges of $q_1=2.6\,\rm nC$ and $q_2=-4.6\,\rm nC$ are $0.12\,\rm m$ apart as shown in the figure below. How much work was done by the electric fields due to those two charges on a charge of $1.5\,\rm nC$ that moves from point $B$ to point $A$?

Solution: the work done by the electric force in moving a charge $q$ between two points with different electric potentials is found by $W=-q\Delta V$, where $\Delta V=V_2-V_1$. In this case, the charge travels from point $B$ to point $A$, so we must first find the potential difference between these two points. The potential at point $A$ is $V_A=V_1+V_2$, where \begin{align*} V_1&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{2.6\times 10^{-9}}{0.13} \\\\&=180\,\rm V \end{align*} and \begin{align*} V_2&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{-4.6\times 10^{-9}}{0.05} \\\\&=-828\,\rm V \end{align*} Therefore, the potential at point $A$ is $\boxed{V_A=180-828=-648\,\rm V}$ Similarly, the potential at point $B$ is $V_B=V_1+V_2$, where \begin{align*} V_1&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{2.6\times 10^{-9}}{0.06}\\\\&=390\,\rm V \end{align*} and \begin{align*} V_2&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{-4.6\times 10^{-9}}{0.06}\\\\&=-690\,\rm V \end{align*} Thus, the potential at point $B$ is $\boxed{V_B=390-690=-300\,\rm V}$ The $1.5-\,\rm nC$ charge moves from $B$ to $A$, so the potential difference between points these points is \begin{align*} \Delta V&=V_{final}-V_{initial} \\&=V_A-V_B \\&=-648-(-300) \\&=\boxed{-148\,\rm V}\end{align*} By having the potential difference between the two points, the work done can be easily obtained as follows \begin{align*} W&=-q\Delta V \\ &=-(1.5\times 10^{-9})(-148) \\&=0.22\times 10^{-6}\,\rm J \end{align*}

Author: Dr. Ali Nemati
Page Published: 2/12/2022