College Physics Quiz: Electric Potential
20 questions · exam conditions
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Electric PotentialQuestion 1 of 20

Two point charges of equal magnitude qq are separated by distance dd. If the electric potential at the midpoint between them is zero, what can be concluded about the charges?

Both charges must be positive
Both charges must be negative
The charges must have opposite signs
One charge must be twice the magnitude of the other
The charges must be located on perpendicular axes
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College Physics Quiz

College Physics Quiz: Electric Potential

Practice Electric Potential in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electric Potential, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two point charges of equal magnitude qq are separated by distance dd. If the electric potential at the midpoint between them is zero, what can be concluded about the charges?

  1. Both charges must be positive
  2. Both charges must be negative
  3. The charges must have opposite signs (correct answer)
  4. One charge must be twice the magnitude of the other
  5. The charges must be located on perpendicular axes
Explanation: When you encounter electric potential problems involving multiple point charges, remember that potential is a scalar quantity that adds algebraically - positive and negative contributions can cancel each other out. The electric potential due to a point charge is V=kqrV = k\frac{q}{r}, where kk is Coulomb's constant, qq is the charge, and rr is the distance from the charge. At the midpoint between two charges separated by distance dd, each charge is at distance d/2d/2 from that point. For the total potential to be zero at the midpoint, the contributions from both charges must cancel: Vtotal=kq1d/2+kq2d/2=0V_{total} = k\frac{q_1}{d/2} + k\frac{q_2}{d/2} = 0. Since the charges have equal magnitude qq, this becomes kqd/2+kqsignd/2=0k\frac{q}{d/2} + k\frac{q_{sign}}{d/2} = 0, where qsignq_{sign} represents the second charge with its sign. For this sum to equal zero, one charge must contribute +kqd/2+k\frac{q}{d/2} while the other contributes −kqd/2-k\frac{q}{d/2}. This only happens when the charges have opposite signs. Answer A is wrong because two positive charges would create positive potential at the midpoint. Answer B is wrong because two negative charges would create negative potential at the midpoint. Answer D is incorrect because the problem states both charges have equal magnitude, and even if they didn't, equal-magnitude charges of the same sign couldn't produce zero potential. Remember: when electric potential equals zero due to multiple charges, look for opposite signs that create canceling contributions.

Question 2

In the region between two parallel plates, the electric field is uniform with magnitude 500 N/C pointing from the positive plate to the negative plate. If the plates are separated by 0.02 m, what is the potential difference between the plates?

  1. 10 V with positive plate at higher potential (correct answer)
  2. 10 V with negative plate at higher potential
  3. 25,000 V with positive plate at higher potential
  4. 25,000 V with negative plate at higher potential
  5. The potential difference cannot be determined without knowing the charge on the plates
Explanation: When you encounter uniform electric fields between parallel plates, you're dealing with one of the most straightforward applications of the relationship between electric field and potential difference. The key equation here is V=EdV = Ed, where V is the potential difference, E is the electric field strength, and d is the distance between the plates. Given an electric field of 500 N/C and plate separation of 0.02 m, the potential difference is: V=(500 N/C)(0.02 m)=10 VV = (500 \text{ N/C})(0.02 \text{ m}) = 10 \text{ V} Since the electric field points from positive to negative plate (as stated), and electric field always points from higher potential to lower potential, the positive plate must be at higher potential than the negative plate. Looking at the wrong answers: Choice B incorrectly suggests the negative plate is at higher potential, which contradicts the fundamental principle that electric field lines point from high to low potential. Choices C and D both give 25,000 V, which results from incorrectly dividing E by d instead of multiplying: 5000.02=25,000\frac{500}{0.02} = 25,000. This is a common algebraic error when students confuse the relationship E=VdE = \frac{V}{d} with the form we need: V=EdV = Ed. Choice A correctly gives 10 V with the positive plate at higher potential. Remember this pattern: for uniform fields between plates, always multiply field strength by distance to get potential difference, and the field direction tells you which plate has higher potential.

Question 3

A uniform electric field of magnitude 1000 N/C points in the positive x-direction. What is the potential difference VB−VAV_B - V_A if point A is at x=0.02x = 0.02 m and point B is at x=0.05x = 0.05 m?

  1. +30+30 V
  2. −30-30 V (correct answer)
  3. +70+70 V
  4. −70-70 V
  5. +50+50 V
Explanation: When you encounter electric field and potential problems, remember that electric potential decreases in the direction of the electric field. The relationship between electric field and potential difference is VB−VA=−∫ABE⃗⋅dl⃗V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}. Since we have a uniform electric field of 1000 N/C pointing in the positive x-direction, and we're moving along the x-axis from point A to point B, the calculation becomes straightforward. The displacement from A to B is Δx=0.05−0.02=0.03\Delta x = 0.05 - 0.02 = 0.03 m in the positive x-direction. For a uniform field, VB−VA=−E⋅Δx=−1000 N/C×0.03 m=−30V_B - V_A = -E \cdot \Delta x = -1000 \text{ N/C} \times 0.03 \text{ m} = -30 V. The negative sign indicates that the potential at B is lower than at A, which makes sense because we're moving in the direction of the electric field. Looking at the wrong answers: Choice A (+30 V) forgot the negative sign in the relationship between field and potential difference. Choice C (+70 V) incorrectly calculated the distance as 0.07 m instead of 0.03 m and also missed the negative sign. Choice D (-70 V) has the correct negative sign but used the wrong distance calculation. Study tip: Always remember that electric potential decreases as you move in the direction of the electric field. The formula VB−VA=−E⋅dV_B - V_A = -E \cdot d for uniform fields automatically gives you the correct sign when you're careful about the direction of movement relative to the field.

Question 4

A parallel-plate capacitor has plates separated by distance dd. If the potential difference between the plates is V0V_0, and a dielectric material with dielectric constant κ\kappa fills half the space between the plates, what is the potential difference across the dielectric-filled region?

  1. V02\frac{V_0}{2}
  2. V0κ+1\frac{V_0}{\kappa + 1} (correct answer)
  3. κV0κ+1\frac{\kappa V_0}{\kappa + 1}
  4. V02κ\frac{V_0}{2\kappa}
  5. V0κ\frac{V_0}{\kappa}
Explanation: When you encounter capacitor problems involving dielectrics, you need to think about how electric fields and potentials distribute across different regions. The key insight is that the electric field strength changes in the dielectric, but the total voltage must add up to the given value. In this configuration, you have two regions in series: one with air (or vacuum) and one filled with dielectric material κ\kappa. Since the same charge appears on both plates, the electric field in the air region is E0E_0 and in the dielectric region is E0/κE_0/\kappa (dielectrics reduce the field strength by factor κ\kappa). Both regions have thickness d/2d/2. The voltage across each region equals the field strength times the distance. If we call the voltage across the dielectric VdV_d and across the air gap VaV_a, then:
  • Va=E0⋅d2V_a = E_0 \cdot \frac{d}{2}
  • Vd=E0κ⋅d2V_d = \frac{E_0}{\kappa} \cdot \frac{d}{2}
Since Va+Vd=V0V_a + V_d = V_0, we get: E0d2(1+1κ)=V0E_0 \frac{d}{2}(1 + \frac{1}{\kappa}) = V_0 Solving for VdV_d: Vd=V0κ+1V_d = \frac{V_0}{\kappa + 1} Answer A (V02\frac{V_0}{2}) incorrectly assumes the voltage splits equally, ignoring the dielectric's effect. Answer C (κV0κ+1\frac{\kappa V_0}{\kappa + 1}) represents the voltage across the air region, not the dielectric. Answer D (V02κ\frac{V_0}{2\kappa}) incorrectly applies the dielectric constant to half the total voltage. Remember: in series configurations with dielectrics, the region with higher dielectric constant gets proportionally less voltage due to its reduced electric field.

Question 5

An alpha particle (charge +2e+2e) is accelerated from rest through a potential difference of 1000 V. What is its final kinetic energy?

  1. 1000 eV
  2. 2000 eV (correct answer)
  3. 500 eV
  4. 4000 eV
  5. 3200 eV
Explanation: When a charged particle moves through a potential difference, it gains kinetic energy equal to the work done by the electric field. This is a fundamental energy conservation principle in electrostatics. To find the kinetic energy gained, use the relationship KE=q⋅ΔVKE = q \cdot \Delta V, where qq is the charge and ΔV\Delta V is the potential difference. An alpha particle has charge +2e+2e, and it's accelerated through 1000 V. Therefore: KE=(2e)(1000 V)=2000 eVKE = (2e)(1000 \text{ V}) = 2000 \text{ eV} This confirms that answer B (2000 eV) is correct. Looking at the wrong answers: Choice A (1000 eV) incorrectly uses q=eq = e instead of q=2eq = 2e—this would be right for a proton but not an alpha particle. Choice C (500 eV) suggests dividing by 2 rather than multiplying, which reverses the relationship between charge and energy gained. Choice D (4000 eV) might result from incorrectly squaring the charge (4e4e instead of 2e2e) or confusing this with kinetic energy formulas like 12mv2\frac{1}{2}mv^2. Remember: when any charged particle accelerates through a potential difference, the energy gained always equals charge times voltage (qΔVq \Delta V). The particle's mass doesn't appear in this calculation—only its charge matters for determining energy gained. This makes these problems straightforward once you identify the particle's charge correctly.

Question 6

A conducting sphere of radius RR carries total charge QQ. What is the electric potential at a point located at distance r=R/2r = R/2 from the center of the sphere?

  1. kQR/2\frac{kQ}{R/2}
  2. kQR\frac{kQ}{R} (correct answer)
  3. 2kQR\frac{2kQ}{R}
  4. Zero
  5. kQ2R\frac{kQ}{2R}
Explanation: When dealing with electrostatic problems involving conducting spheres, you need to understand how charge distributes and how potential behaves both inside and outside the conductor. For any conductor in electrostatic equilibrium, all excess charge resides on the surface, and the electric field inside is zero. Since electric potential is related to the electric field, the potential is constant throughout the entire interior of the conductor, including at its surface. To find this constant potential, we calculate the potential at the surface (where r=Rr = R). Using the formula for potential due to a point charge, V=kQrV = \frac{kQ}{r}, the potential at the surface is V=kQRV = \frac{kQ}{R}. Since potential is constant everywhere inside the conductor, the potential at r=R/2r = R/2 is also kQR\frac{kQ}{R}, making choice B correct. Let's examine why the other options are wrong: Choice A (kQR/2\frac{kQ}{R/2}) incorrectly applies the point charge formula directly to the interior point, ignoring that we're inside a conductor. Choice C (2kQR\frac{2kQ}{R}) makes a similar error, perhaps from algebraic manipulation of choice A. Choice D (Zero) confuses electric field with electric potential—while the field is zero inside, the potential is not. Study tip: Remember that inside any conductor, the electric field is always zero, but the electric potential is constant and non-zero (unless the conductor is grounded). Always check whether you're asked about a point inside or outside the conductor, as this completely changes your approach.

Question 7

Two parallel conducting plates create a uniform electric field. An electron released from rest at the negative plate reaches the positive plate with speed vv. If the distance between plates is doubled while keeping the same potential difference, what speed will the electron have when it reaches the positive plate?

  1. v2\frac{v}{2}
  2. v2\frac{v}{\sqrt{2}}
  3. vv (correct answer)
  4. v2v\sqrt{2}
  5. 2v2v
Explanation: When analyzing motion in electric fields, you need to focus on energy conservation and how electric field strength relates to plate geometry. Initially, the electron gains kinetic energy equal to the work done by the electric field: 12mv2=eV\frac{1}{2}mv^2 = eV, where VV is the potential difference. This establishes the relationship between the electron's final speed and the voltage. When you double the distance while keeping the same potential difference, the electric field strength becomes E=V2dE = \frac{V}{2d} instead of E=VdE = \frac{V}{d}. However, the crucial insight is that the work done on the electron depends only on the potential difference: W=eVW = eV. Since the potential difference remains unchanged, the electron still gains the same kinetic energy: 12mv2=eV\frac{1}{2}mv^2 = eV. Therefore, the final speed remains vv. Choice A (v2\frac{v}{2}) incorrectly assumes the speed is inversely proportional to distance. Choice B (v2\frac{v}{\sqrt{2}}) mistakenly applies the relationship E∝1dE \propto \frac{1}{d} directly to kinetic energy, forgetting that energy depends on voltage, not field strength. Choice D (v2v\sqrt{2}) wrongly suggests the electron gains more energy with increased distance. The correct answer is C. Key strategy: In electric field problems, always distinguish between field strength (which depends on geometry) and potential difference (which determines energy changes). Work done by conservative electric forces depends only on the potential difference between starting and ending points, regardless of the path or field configuration.

Question 8

An electric dipole consists of charges +q+q and −q-q separated by distance dd. At a point on the perpendicular bisector of the dipole, far from the dipole (r≫dr \gg d), the electric potential is approximately:

  1. kqdr2\frac{kqd}{r^2}
  2. kqr\frac{kq}{r}
  3. kqd2r3\frac{kqd^2}{r^3}
  4. Zero (correct answer)
  5. 2kqr\frac{2kq}{r}
Explanation: When analyzing electric dipole problems, you need to consider how the contributions from both charges combine at the point of interest. An electric dipole creates different field patterns depending on your location relative to the charges. The electric potential at any point is the scalar sum of potentials from both charges: V=kqr++k(−q)r−=kq(1r+−1r−)V = \frac{kq}{r_+} + \frac{k(-q)}{r_-} = kq\left(\frac{1}{r_+} - \frac{1}{r_-}\right), where r+r_+ and r−r_- are distances from the positive and negative charges respectively. On the perpendicular bisector, you're equidistant from both charges, so r+=r−=rr_+ = r_- = r. This means the potential becomes V=kq(1r−1r)=0V = kq\left(\frac{1}{r} - \frac{1}{r}\right) = 0. The contributions from the positive and negative charges exactly cancel out. Choice A (kqdr2\frac{kqd}{r^2}) represents the leading term in the electric field magnitude along the dipole axis, not the potential on the perpendicular bisector. Choice B (kqr\frac{kq}{r}) would be the potential from a single point charge, ignoring the negative charge entirely. Choice C (kqd2r3\frac{kqd^2}{r^3}) has the wrong distance dependence and includes d2d^2 rather than the linear dd dependence characteristic of dipole moments. Remember this key insight: on the perpendicular bisector of any dipole, the potential is always zero due to symmetry. This is true regardless of distance, making perpendicular bisector problems straightforward once you recognize the geometric relationship.

Question 9

A metal sphere of radius 0.10 m is given a charge of 3.0×10−63.0 \times 10^{-6} C. If the sphere is then connected to a second uncharged metal sphere of radius 0.20 m by a conducting wire, what is the final potential of both spheres?

  1. 9.0×1049.0 \times 10^4 V (correct answer)
  2. 1.35×1051.35 \times 10^5 V
  3. 2.7×1052.7 \times 10^5 V
  4. 4.5×1044.5 \times 10^4 V
  5. 1.8×1051.8 \times 10^5 V
Explanation: When two conducting spheres are connected by a wire, they reach the same electric potential - this is a fundamental principle of electrostatics. The key insight is that charge redistributes between the spheres until equilibrium is achieved. Initially, the smaller sphere (radius 0.10 m) has all the charge: Q1=3.0×10−6Q_1 = 3.0 \times 10^{-6} C. When connected, charge flows until both spheres have identical potential: V1=V2V_1 = V_2. For a conducting sphere, V=kQrV = k\frac{Q}{r}, so at equilibrium: kQ1r1=kQ2r2k\frac{Q_1}{r_1} = k\frac{Q_2}{r_2} This gives us: Q10.10=Q20.20\frac{Q_1}{0.10} = \frac{Q_2}{0.20}, or Q2=2Q1Q_2 = 2Q_1 Since total charge is conserved: Q1+Q2=3.0×10−6Q_1 + Q_2 = 3.0 \times 10^{-6} C Substituting: Q1+2Q1=3.0×10−6Q_1 + 2Q_1 = 3.0 \times 10^{-6}, so Q1=1.0×10−6Q_1 = 1.0 \times 10^{-6} C The final potential is: V=kQ1r1=(9.0×109)1.0×10−60.10=9.0×104V = k\frac{Q_1}{r_1} = (9.0 \times 10^9) \frac{1.0 \times 10^{-6}}{0.10} = 9.0 \times 10^4 V This confirms answer A is correct. Answer B (1.35×1051.35 \times 10^5 V) incorrectly assumes charge distributes proportionally to surface area. Answer C (2.7×1052.7 \times 10^5 V) represents the initial potential of the small sphere before connection. Answer D (4.5×1044.5 \times 10^4 V) results from incorrectly averaging the initial potentials. Remember: connected conductors always reach the same potential, with charge redistributing proportionally to radius, not surface area or volume.

Question 10

A point charge +Q+Q is located at the origin. The electric potential at point P, located at coordinates (3, 4) (in meters), is 180 V. What is the electric potential at point R, located at coordinates (6, 8)?

  1. 45 V
  2. 90 V (correct answer)
  3. 180 V
  4. 360 V
  5. 720 V
Explanation: When you encounter electric potential problems with point charges, remember that potential depends only on the distance from the charge, not the specific coordinates. The electric potential from a point charge follows V=kQrV = \frac{kQ}{r}, where rr is the distance from the charge. First, find the distance from the origin to point P at (3, 4): rP=32+42=9+16=5 mr_P = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \text{ m}. Since VP=180 VV_P = 180 \text{ V}, you can write: 180=kQ5180 = \frac{kQ}{5}, so kQ=900kQ = 900. Next, find the distance to point R at (6, 8): rR=62+82=36+64=10 mr_R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10 \text{ m}. Notice that point R is exactly twice as far from the origin as point P. Since potential is inversely proportional to distance, VR=kQ10=90010=90 VV_R = \frac{kQ}{10} = \frac{900}{10} = 90 \text{ V}. Answer choice A (45 V) incorrectly assumes potential decreases as the square of distance, like electric field intensity. Answer choice C (180 V) wrongly suggests potential is the same at both points, ignoring the distance dependence. Answer choice D (360 V) mistakenly applies a direct proportionality, as if potential increases with distance. The correct answer is B (90 V). Study tip: Always calculate distances using the Pythagorean theorem for coordinate problems, and remember that electric potential from a point charge has a simple 1/r1/r dependence—if you double the distance, you halve the potential.

Question 11

Two conducting spheres of radii R1=0.1R_1 = 0.1 m and R2=0.2R_2 = 0.2 m are connected by a conducting wire. If the total charge on the system is Q=6.0×10−6Q = 6.0 \times 10^{-6} C, what is the potential of the system?

  1. 1.8×105 V1.8 \times 10^5 \text{ V} (correct answer)
  2. 2.7×105 V2.7 \times 10^5 \text{ V}
  3. 5.4×105 V5.4 \times 10^5 \text{ V}
  4. 9.0×105 V9.0 \times 10^5 \text{ V}
  5. 1.08×106 V1.08 \times 10^6 \text{ V}
Explanation: When you encounter conducting spheres connected by a wire, remember that they must be at the same electric potential since they're conductively connected. This is a key principle: conductors in contact share the same potential. Since both spheres are at the same potential VV, you can write: V=kQ1R1=kQ2R2V = k\frac{Q_1}{R_1} = k\frac{Q_2}{R_2}, where Q1Q_1 and Q2Q_2 are the charges on each sphere. This gives us Q1R1=Q2R2\frac{Q_1}{R_1} = \frac{Q_2}{R_2}, so Q1=Q2R1R2=Q20.10.2=Q22Q_1 = Q_2 \frac{R_1}{R_2} = Q_2 \frac{0.1}{0.2} = \frac{Q_2}{2}. Since the total charge is conserved: Q1+Q2=6.0×10−6Q_1 + Q_2 = 6.0 \times 10^{-6} C. Substituting: Q22+Q2=6.0×10−6\frac{Q_2}{2} + Q_2 = 6.0 \times 10^{-6}, which gives Q2=4.0×10−6Q_2 = 4.0 \times 10^{-6} C and Q1=2.0×10−6Q_1 = 2.0 \times 10^{-6} C. The potential is: V=kQ1R1=(9.0×109)2.0×10−60.1=1.8×105V = k\frac{Q_1}{R_1} = (9.0 \times 10^9) \frac{2.0 \times 10^{-6}}{0.1} = 1.8 \times 10^5 V. Answer A (1.8×1051.8 \times 10^5 V) is correct. Answer B (2.7×1052.7 \times 10^5 V) likely comes from incorrect charge distribution assumptions. Answer C (5.4×1055.4 \times 10^5 V) might result from using total charge with one radius incorrectly. Answer D (9.0×1059.0 \times 10^5 V) suggests using the wrong radius or misapplying the total charge. Key strategy: Always remember that connected conductors have equal potential, and charge distributes proportionally to radius. Set up the equal potential condition first, then solve for individual charges.

Question 12

A positive point charge +Q+Q is fixed at the origin. A second positive point charge +q+q is moved from infinity to a point at distance rr from the origin. Which statement correctly describes the work done and potential energy change?

  1. Positive work is done by the external force, and the potential energy increases (correct answer)
  2. Negative work is done by the external force, and the potential energy decreases
  3. Positive work is done by the external force, and the potential energy decreases
  4. No net work is done, and the potential energy remains constant
  5. The work done depends on the path taken, but potential energy change is always positive
Explanation: When you encounter electrostatics problems involving moving charges, focus on the relationship between conservative forces, work, and potential energy. The key insight is that electric potential energy depends on the configuration of charges, and work must be done against repulsive forces. Since both charges are positive, they repel each other. As you bring charge +q+q from infinity (where potential energy is zero) to distance rr, you're moving it against the repulsive electric force from +Q+Q. This requires an external agent to do positive work. The potential energy of the system increases because U=kQqrU = k\frac{Qq}{r} becomes positive and larger as rr decreases from infinity to a finite value. Choice A correctly identifies both aspects: positive work by the external force and increased potential energy. Choice B incorrectly suggests negative external work and decreased potential energy. This would occur if you were moving the charge away from the fixed charge, not toward it. Choice C wrongly claims potential energy decreases while external work is positive. This violates energy conservation - when you do positive work on a system with conservative forces, the potential energy must increase. Choice D incorrectly states no net work is done. While the net work by all forces on a charge moved at constant velocity is zero, the question specifically asks about work done by the external force, which must be positive to overcome the repulsive electric force. Remember: moving like charges together always requires positive external work and increases electric potential energy. The external work equals the change in potential energy.

Question 13

In a certain region, the electric potential increases linearly from 0 V at x=0x = 0 to 100 V at x=0.5x = 0.5 m. A proton moves from x=0.1x = 0.1 m to x=0.4x = 0.4 m. What is the change in the proton's electric potential energy?

  1. +60+60 eV (correct answer)
  2. −60-60 eV
  3. +80+80 eV
  4. −80-80 eV
  5. +20+20 eV
Explanation: When you encounter electric potential problems, remember that potential energy depends on both the charge and the electric potential at each location. The key relationship is U=qVU = qV, where UU is potential energy, qq is charge, and VV is electric potential. First, you need to find the electric potential at each position. Since the potential increases linearly from 0 V at x=0x = 0 to 100 V at x=0.5x = 0.5 m, the potential varies as V(x)=200xV(x) = 200x volts. At x=0.1x = 0.1 m, V=20V = 20 V. At x=0.4x = 0.4 m, V=80V = 80 V. The change in potential energy is ΔU=q(Vf−Vi)=q(80−20)=60q\Delta U = q(V_f - V_i) = q(80 - 20) = 60q volts. Since a proton has charge +e+e, the change is +60e=+60+60e = +60 eV. This makes answer A correct. Let's examine why the other options are wrong. Answer B (−60-60 eV) incorrectly uses a negative sign, which would apply if the proton moved from high to low potential instead of low to high. Answer C (+80+80 eV) mistakenly uses only the final potential (80 V) rather than the change in potential (60 V). Answer D (−80-80 eV) combines both errors: using only the final potential and applying an incorrect negative sign. Remember this pattern: when a positive charge moves to higher potential, its potential energy increases (positive ΔU\Delta U). Always calculate the change in potential first, then multiply by the charge to find the energy change.

Question 14

A proton moves from point A where the electric potential is 100 V to point B where the electric potential is 40 V. What is the change in the proton's electric potential energy?

  1. −60 eV-60 \text{ eV} (correct answer)
  2. −9.6×10−18 J-9.6 \times 10^{-18} \text{ J}
  3. +60 eV+60 \text{ eV}
  4. +9.6×10−18 J+9.6 \times 10^{-18} \text{ J}
  5. +140 eV+140 \text{ eV}
Explanation: When you encounter electric potential energy problems, remember that potential energy depends on both the charge and the electric potential at each location. The key relationship is ΔU=qΔV\Delta U = q \Delta V, where ΔU\Delta U is the change in potential energy, qq is the charge, and ΔV\Delta V is the change in electric potential. For this proton moving from 100 V to 40 V, the change in potential is ΔV=VB−VA=40 V−100 V=−60 V\Delta V = V_B - V_A = 40\text{ V} - 100\text{ V} = -60\text{ V}. Since a proton has a positive charge of +e+e (one elementary charge), the change in potential energy is: ΔU=qΔV=(+e)(−60 V)=−60 eV\Delta U = q \Delta V = (+e)(-60\text{ V}) = -60\text{ eV} This makes answer A correct. The negative sign indicates the proton loses potential energy as it moves from higher to lower potential. Answer B gives the correct magnitude but in joules rather than electron volts, missing that energy problems involving single particles are typically expressed in eV. Answer C has the wrong sign—this would suggest the proton gained energy moving to lower potential, which contradicts the physics. Answer D combines both errors: wrong units and wrong sign. The key insight is that positive charges naturally move from high to low potential, losing potential energy in the process, just like a ball rolling downhill loses gravitational potential energy. Study tip: Always check the sign carefully in potential energy problems. Positive charges lose potential energy when moving to lower potential, while negative charges gain potential energy in the same situation.

Question 15

Two point charges, +3.0 μC+3.0 \text{ μC} and −2.0 μC-2.0 \text{ μC}, are separated by a distance of 0.60 m0.60 \text{ m}. What is the electric potential at a point that is 0.30 m0.30 \text{ m} from the positive charge and 0.30 m0.30 \text{ m} from the negative charge?

  1. 9.0×104 V9.0 \times 10^4 \text{ V}
  2. 3.0×104 V3.0 \times 10^4 \text{ V} (correct answer)
  3. 1.5×104 V1.5 \times 10^4 \text{ V}
  4. 4.5×104 V4.5 \times 10^4 \text{ V}
Explanation: The electric potential at a point due to multiple point charges is the scalar sum of the potentials due to each charge: V=kq1r1+kq2r2V = k\frac{q_1}{r_1} + k\frac{q_2}{r_2}. Here, V=(9.0×109)(3.0×10−60.30+−2.0×10−60.30)=(9.0×109)(1.0×10−60.30)=3.0×104 VV = (9.0 \times 10^9)\left(\frac{3.0 \times 10^{-6}}{0.30} + \frac{-2.0 \times 10^{-6}}{0.30}\right) = (9.0 \times 10^9)\left(\frac{1.0 \times 10^{-6}}{0.30}\right) = 3.0 \times 10^4 \text{ V}. Choice A incorrectly adds the magnitudes without considering the negative charge's negative contribution. Choice C uses only half the net charge. Choice D incorrectly multiplies by 1.5 instead of adding algebraically.

Question 16

A conducting sphere of radius RR carries a total charge QQ. What is the electric potential at a distance r=R/2r = R/2 from the center of the sphere?

  1. kQR/2=2kQR\frac{kQ}{R/2} = \frac{2kQ}{R}
  2. Zero, because the point is inside the conductor
  3. kQ2R\frac{kQ}{2R}
  4. kQR\frac{kQ}{R} (correct answer)
Explanation: When dealing with conducting spheres, you need to understand how charge distributes and how electric potential behaves both inside and outside conductors. In a conductor at electrostatic equilibrium, all charges reside on the surface, and the electric field inside is zero. However, this doesn't mean the electric potential inside is zero. Electric potential is related to the work needed to bring a charge from infinity, and inside a conductor, the potential is constant and equal to the potential at the surface. To find the potential at r=R/2r = R/2 (inside the sphere), we use the fact that potential is constant throughout the conductor. The potential at any point inside equals the surface potential: V=kQRV = \frac{kQ}{R}. Since the point r=R/2r = R/2 is inside the conductor, the potential there is kQR\frac{kQ}{R}, making D correct. Looking at the wrong answers: A applies the potential formula V=kQrV = \frac{kQ}{r} directly using r=R/2r = R/2, which would be correct for a point charge but ignores that we're inside a conductor where potential is constant. B incorrectly assumes that zero electric field inside means zero potential—this confuses field with potential. C appears to be kQR\frac{kQ}{R} divided by 2, perhaps incorrectly thinking the potential scales with the distance ratio. Remember: inside any conductor, electric potential is constant and equals the surface potential. Don't confuse this with the electric field, which is zero inside but doesn't determine the potential value.

Question 17

Two identical parallel plates separated by distance dd have a potential difference of V0V_0. A dielectric material with dielectric constant κ=3.0\kappa = 3.0 completely fills the space between the plates. If the plates remain connected to the battery maintaining constant voltage, what is the new electric potential energy stored in the capacitor compared to the original energy without the dielectric?

  1. The energy increases by a factor of 3.0 because the capacitance increases by this factor. (correct answer)
  2. The energy decreases by a factor of 3.0 because the electric field decreases by this factor.
  3. The energy increases by a factor of 3.0 because both charge and capacitance increase proportionally.
  4. The energy remains unchanged because the voltage across the plates stays constant.
Explanation: When a dielectric is inserted while maintaining constant voltage, the capacitance increases by a factor of κ\kappa: C′=κCC' = \kappa C. The energy stored is U=12CV2U = \frac{1}{2}CV^2. Since voltage is constant, U′=12κCV2=κUU' = \frac{1}{2}\kappa C V^2 = \kappa U. The energy increases by a factor of 3.0. Choice B incorrectly focuses on the field decrease and gets the wrong direction of energy change. Choice C has correct reasoning about capacitance but incorrectly mentions charge increasing 'proportionally' when charge increases by factor κ\kappa, not proportionally to capacitance and voltage together. Choice D incorrectly assumes energy depends only on voltage.

Question 18

A uniform electric field E⃗=200 N/C\vec{E} = 200 \text{ N/C} (in the +x+x direction) exists in a region. Point A is at the origin and point B is at coordinates (0.05 m,0.03 m)(0.05 \text{ m}, 0.03 \text{ m}). What is the potential difference VB−VAV_B - V_A?

  1. +6 V+6 \text{ V}
  2. −16 V-16 \text{ V}
  3. +10 V+10 \text{ V}
  4. −10 V-10 \text{ V} (correct answer)
Explanation: When you encounter electric field and potential problems, remember that electric potential decreases in the direction of the electric field. The relationship between electric field and potential difference is VB−VA=−E⃗⋅Δr⃗V_B - V_A = -\vec{E} \cdot \Delta\vec{r}, where Δr⃗\Delta\vec{r} is the displacement vector from A to B. To find the potential difference, you need the displacement vector from point A (origin) to point B. This gives Δr⃗=(0.05,0.03) m\Delta\vec{r} = (0.05, 0.03) \text{ m}. The electric field vector is E⃗=(200,0) N/C\vec{E} = (200, 0) \text{ N/C} since it points in the +x direction. The dot product E⃗⋅Δr⃗=200×0.05+0×0.03=10 V\vec{E} \cdot \Delta\vec{r} = 200 \times 0.05 + 0 \times 0.03 = 10 \text{ V}. Therefore, VB−VA=−10 VV_B - V_A = -10 \text{ V}, which is answer D. Let's examine why the other options are wrong: Option A (+6 V) appears to use incorrect coordinates or calculation. Option B (-16 V) might result from incorrectly using the total distance 0.052+0.032=0.058 m\sqrt{0.05^2 + 0.03^2} = 0.058 \text{ m} instead of just the x-component displacement. Option C (+10 V) gets the magnitude right but misses the crucial negative sign—this is the most common error, forgetting that potential decreases in the field direction. The key strategy here is remembering that only the component of displacement parallel to the electric field matters for potential difference calculations. The y-component displacement is irrelevant since the field has no y-component. Always pay attention to the negative sign in the potential-field relationship.

Question 19

Two concentric conducting spheres have radii R1=2.0 cmR_1 = 2.0 \text{ cm} and R2=4.0 cmR_2 = 4.0 \text{ cm}. The inner sphere carries charge +Q+Q and the outer sphere carries charge −Q-Q. What is the electric potential at a point located at r=3.0 cmr = 3.0 \text{ cm} from the center (between the spheres)?

  1. kQ0.03−kQ0.04\frac{kQ}{0.03} - \frac{kQ}{0.04}
  2. kQ0.03\frac{kQ}{0.03}
  3. kQ0.02−kQ0.04\frac{kQ}{0.02} - \frac{kQ}{0.04} (correct answer)
  4. Zero, because the charges are equal and opposite
Explanation: When dealing with electric potential from multiple charged conductors, you need to apply the principle of superposition - the total potential at any point is the sum of potentials from each charge source. For this configuration, the potential at r=3.0 cm=0.03 mr = 3.0 \text{ cm} = 0.03 \text{ m} comes from two sources: the inner sphere with charge +Q+Q and the outer sphere with charge −Q-Q. Since we're outside the inner sphere (r>R1r > R_1), its charge acts as if concentrated at the center, giving potential kQ0.03\frac{kQ}{0.03}. Since we're inside the outer sphere (r<R2r < R_2), we use the outer sphere's surface potential, which is −kQR2=−kQ0.04-\frac{kQ}{R_2} = -\frac{kQ}{0.04}. The total potential is kQ0.02−kQ0.04\frac{kQ}{0.02} - \frac{kQ}{0.04}, making answer C correct. Answer A incorrectly uses r=0.03r = 0.03 for both terms, failing to recognize that the outer sphere's potential at any interior point equals its surface potential. Answer B only accounts for the inner sphere's contribution, completely ignoring the outer sphere. Answer D falls into the common misconception that equal and opposite charges always cancel - while the net charge is zero, the potential contributions don't cancel because the charges are at different distances from our point of interest. Remember: for conductors, always check whether your point is inside or outside each conductor. Inside a conductor, use the surface potential; outside, treat the charge as point-like at the center.

Question 20

A uniform electric field of magnitude 500 N/C500 \text{ N/C} points in the positive xx-direction. An electron moves from point A at (2.0 cm,0)(2.0 \text{ cm}, 0) to point B at (8.0 cm,3.0 cm)(8.0 \text{ cm}, 3.0 \text{ cm}). What is the change in electric potential energy of the electron during this displacement?

  1. −4.8×10−18 J-4.8 \times 10^{-18} \text{ J}
  2. +4.8×10−18 J+4.8 \times 10^{-18} \text{ J} (correct answer)
  3. −7.2×10−18 J-7.2 \times 10^{-18} \text{ J}
  4. +7.2×10−18 J+7.2 \times 10^{-18} \text{ J}
Explanation: The change in potential energy is ΔU=qΔV=q(−EΔx)\Delta U = q\Delta V = q(-E\Delta x) since the field points in the +x+x direction. The displacement in the xx-direction is Δx=8.0−2.0=6.0 cm=0.060 m\Delta x = 8.0 - 2.0 = 6.0 \text{ cm} = 0.060 \text{ m}. For an electron, q=−1.6×10−19 Cq = -1.6 \times 10^{-19} \text{ C}. So ΔU=(−1.6×10−19)(−500)(0.060)=+4.8×10−18 J\Delta U = (-1.6 \times 10^{-19})(-500)(0.060) = +4.8 \times 10^{-18} \text{ J}. The electron gains potential energy as it moves against the electric force. Choice A has the wrong sign. Choices C and D incorrectly include the yy-displacement, which doesn't affect the potential in a uniform field pointing in the xx-direction.