Electric Potential Energy and Potential — Hard Practice Quiz

A Physics cheat sheet for Electric Potential Energy and Potential — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Formulas & key concepts

Work and Potential Energy: Work \(W_{AB}\) done by electric force as charge moves from A to B equals change in electric potential energy \(EPE\).

$$W_{AB} = EPE_A - EPE_B$$

Electric Potential Definition: Electric potential \(V\) is the electric potential energy \(EPE\) per unit charge \(q_0\).

$$V = \frac{EPE}{q_0}$$

Potential Difference: Difference in potential between points B and A is related to work done by electric force.

$$V_B - V_A = \frac{-W_{AB}}{q_0}$$

Potential of a Point Charge: Electric potential \(V\) at distance \(r\) from a point charge \(q\).

$$V = k \frac{q}{r}$$

Capacitor Charge: Charge \(q\) on a capacitor is proportional to potential difference \(V\) across it. \(C\) is capacitance.

$$q = CV$$

Parallel Plate Capacitor: Capacitance \(C\) depends on area \(A\), separation \(d\), and dielectric constant \(\kappa\).

$$C = \frac{\kappa \epsilon_0 A}{d}$$

Where: \(\epsilon_0\) = permittivity of free space

Energy Stored in Capacitor: Potential energy stored in a charged capacitor.

$$Energy = \frac{1}{2}CV^2 = \frac{1}{2}qV = \frac{q^2}{2C}$$

Practice quiz

  1. A point charge $Q$ is located at the origin. A test charge $q_0$ is moved from infinity to a distance $r$ from $Q$. What is the electric potential energy $EPE$ of the test charge at distance $r$?

    • $k \frac{Q}{r}$
    • $k \frac{q_0}{r}$
    • $k \frac{Q q_0}{r}$
    • $k \frac{Q q_0}{r^2}$

    Answer: $k \frac{Q q_0}{r}$

  2. A charge $q_0$ moves from point A to point B in an electric field. The electric potential at point A is $V_A$ and at point B is $V_B$. If $W_{AB}$ is the work done by the electric field on the charge, which of the following relationships is correct?

    • $W_{AB} = q_0 (V_B - V_A)$
    • $W_{AB} = q_0 (V_A - V_B)$
    • $W_{AB} = \frac{V_A - V_B}{q_0}$
    • $W_{AB} = \frac{V_B - V_A}{q_0}$

    Answer: $W_{AB} = q_0 (V_A - V_B)$

  3. A parallel plate capacitor with capacitance $C_0$ is charged to a potential difference $V_0$ and then disconnected from the battery. The initial energy stored is $Energy_0$. If the plate separation $d$ is doubled and a dielectric with constant $\kappa = 2$ is inserted, what is the new energy stored in the capacitor?

    • $4 Energy_0$
    • $2 Energy_0$
    • $Energy_0$
    • $\frac{1}{2} Energy_0$

    Answer: $Energy_0$

  4. A capacitor is charged by a battery to a potential $V$. If the capacitance is doubled while keeping the charge on the plates constant, how does the stored energy change?

    • It becomes four times the original energy.
    • It becomes twice the original energy.
    • It remains the same.
    • It becomes half the original energy.

    Answer: It becomes half the original energy.

  5. Two points A and B are at distances $r_A$ and $r_B$ respectively from a point charge $Q$. If a test charge $q_0$ moves from A to B, what is the work done by the electric field $W_{AB}$ in terms of $Q$, $q_0$, $r_A$, $r_B$, and $k$?

    • $k Q q_0 (\frac{1}{r_B} - \frac{1}{r_A})$
    • $k Q q_0 (\frac{1}{r_A} - \frac{1}{r_B})$
    • $k \frac{Q q_0}{r_A r_B}$
    • $k \frac{Q q_0}{(r_A - r_B)^2}$

    Answer: $k Q q_0 (\frac{1}{r_A} - \frac{1}{r_B})$

  6. A parallel plate capacitor is connected to a battery, establishing a potential difference $V$. If the plate area $A$ is doubled and the plate separation $d$ is halved, how does the stored energy change?

    • It becomes four times the original energy.
    • It becomes twice the original energy.
    • It remains the same.
    • It becomes half the original energy.

    Answer: It becomes four times the original energy.

  7. A positive point charge $Q$ is fixed at the origin. A small positive test charge $q_0$ is moved from point A at $r_A$ to point B at $r_B$, where $r_B > r_A$. Which of the following statements is true regarding the work done by the electric field $W_{AB}$ and the change in electric potential energy $EPE_B - EPE_A$?

    • $W_{AB}$ is positive, and $EPE_B - EPE_A$ is positive.
    • $W_{AB}$ is negative, and $EPE_B - EPE_A$ is negative.
    • $W_{AB}$ is positive, and $EPE_B - EPE_A$ is negative.
    • $W_{AB}$ is negative, and $EPE_B - EPE_A$ is positive.

    Answer: $W_{AB}$ is positive, and $EPE_B - EPE_A$ is negative.

  8. Express the potential difference $V$ across a capacitor in terms of its stored energy $Energy$ and charge $q$.

    • $V = \frac{Energy}{q}$
    • $V = \frac{2 Energy}{q}$
    • $V = \frac{q}{2 Energy}$
    • $V = \frac{q Energy}{2}$

    Answer: $V = \frac{2 Energy}{q}$

  9. A parallel plate capacitor with air ($\kappa=1$) between its plates has capacitance $C_0$. It is charged to a potential $V_0$ and then disconnected from the battery. A dielectric slab with constant $\kappa=3$ is then inserted, filling the space between the plates. What is the ratio of the new potential difference $V_f$ to the original potential difference $V_0$?

    • $3$
    • $1$
    • $\frac{1}{3}$
    • $\frac{1}{9}$

    Answer: $\frac{1}{3}$

  10. Two point charges, $q_1 = +2q$ and $q_2 = -q$, are placed at $(0,0)$ and $(a,0)$ respectively. What is the electric potential energy of a third point charge $q_3 = +q$ placed at $(0,a)$?

    • $k \frac{q^2}{a} (2 + \frac{1}{\sqrt{2}})$
    • $k \frac{q^2}{a} (2 - \frac{1}{\sqrt{2}})$
    • $k \frac{q^2}{a} (\frac{1}{\sqrt{2}} - 2)$
    • $k \frac{q^2}{a} (2 - \sqrt{2})$

    Answer: $k \frac{q^2}{a} (2 - \frac{1}{\sqrt{2}})$

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