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 \(EPE\).
Electric Potential Definition: Electric potential \(V\) is the electric potential energy \(EPE\) per unit charge \(q_0\).
Potential Difference: Difference in potential between points B and A is related to work done by electric force.
Potential of a Point Charge: Electric potential \(V\) at distance \(r\) from a point charge \(q\).
Capacitor Charge: Charge \(q\) on a capacitor is proportional to potential difference \(V\) across it. \(C\) is capacitance.
Parallel Plate Capacitor: Capacitance \(C\) depends on area \(A\), separation \(d\), and dielectric constant \(\kappa\).
Where: \(\epsilon_0\) = permittivity of free space
Energy Stored in Capacitor: Potential energy stored in a charged capacitor.
Practice quiz
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: $k \frac{Q q_0}{r}$
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: $W_{AB} = q_0 (V_A - V_B)$
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: $Energy_0$
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.
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: $k Q q_0 (\frac{1}{r_A} - \frac{1}{r_B})$
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.
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.
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: $V = \frac{2 Energy}{q}$
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: $\frac{1}{3}$
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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