Electric Potential & Potential Energy — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Electric Potential & Potential Energy MCQs with step-by-step solutions (21 questions). Part of Electrostatics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Electric Potential & Potential Energy · easy · theory
The electric potential at a point is defined as:
A. The charge per unit area
B. The energy stored per unit volume
C. The force per unit charge at that point
D. The work done per unit positive charge in bringing it from infinity to that point ✓ Correct
Solution: Its SI unit is the volt, equal to one joule per coulomb.
Q2 — Electric Potential & Potential Energy · easy · theory
The electric potential at distance $r$ from a point charge $q$ is:
A. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$
B. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ ✓ Correct
C. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r}$
D. $\dfrac{1}{4\pi\varepsilon_0}qr$
Solution: Unlike the field, the potential falls off as $\dfrac{1}{r}$ and is a scalar, so potentials add algebraically.
Q3 — Electric Potential & Potential Energy · medium · theory
The electric potential everywhere inside a charged conducting shell is:
A. Greater than the surface potential
B. Infinite at the centre
C. Zero
D. Equal to the potential on its surface ✓ Correct
Solution: Since the interior field is zero, no work is needed to move a charge about inside, so the potential is uniform.
Q4 — Electric Potential & Potential Energy · medium · theory
The relation between electric field and potential is:
A. $E = +\dfrac{dV}{dr}$
B. $E = \dfrac{V}{r^2}$
C. $E = -\dfrac{dV}{dr}$ ✓ Correct
D. $E = -V r$
Solution: The field points in the direction of steepest decrease of potential, which the minus sign expresses.
Q5 — Electric Potential & Potential Energy · medium · theory
Equipotential surfaces are always:
A. Parallel to the electric lines of force
B. Spherical in every situation
C. Inclined at $45^\circ$ to the field
D. Perpendicular to the electric lines of force ✓ Correct
Solution: If the field had a component along the surface, work would be needed to move a charge over it and the surface would not be equipotential.
Q6 — Electric Potential & Potential Energy · easy · theory
The work done in moving a charge between two points on the same equipotential surface is:
A. Zero ✓ Correct
B. Equal to $qV$
C. Maximum
D. Negative
Solution: $W = q(V_2 - V_1) = 0$ because the potential difference vanishes.
Q7 — Electric Potential & Potential Energy · easy · theory
Electric potential is a:
A. Vector quantity
B. Tensor quantity
C. Dimensionless quantity
D. Scalar quantity ✓ Correct
Solution: Potentials due to several charges add as ordinary numbers, which often makes potential easier to compute than field.
Q8 — Electric Potential & Potential Energy · medium · theory
The work done in moving a test charge from the surface of a charged hollow metallic sphere to its centre is:
A. $\dfrac{q_0Q}{8\pi\varepsilon_0R}$
B. $\dfrac{q_0Q}{4\pi\varepsilon_0R}$
C. Zero ✓ Correct
D. Infinite
Solution: The interior is an equipotential region because $E = 0$, so no work is required.
Q9 — Electric Potential & Potential Energy · medium · numerical
The potential at $0.1\text{ m}$ from a point charge of $1\,\mu\text{C}$ is:
A. $9 \times 10^5\text{ V}$
B. $9 \times 10^3\text{ V}$
C. $10^5\text{ V}$
D. $9 \times 10^4\text{ V}$ ✓ Correct
Solution: $V = \dfrac{kq}{r} = \dfrac{9 \times 10^9 \times 10^{-6}}{0.1} = 9 \times 10^4\text{ V}$.
Q10 — Electric Potential & Potential Energy · medium · numerical
The potential at $0.3\text{ m}$ from a charge of $2\,\mu\text{C}$ is:
A. $1.8 \times 10^4\text{ V}$
B. $6 \times 10^4\text{ V}$ ✓ Correct
C. $6 \times 10^5\text{ V}$
D. $2 \times 10^5\text{ V}$
Solution: $V = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.3} = \dfrac{1.8 \times 10^4}{0.3} = 6 \times 10^4\text{ V}$.
Q11 — Electric Potential & Potential Energy · hard · numerical
A metal shell of radius $R$ carries charge $Q$. The potential at a distance $\dfrac{R}{2}$ from its centre is:
A. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Q}{R}$
B. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$ ✓ Correct
C. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$
D. Zero
Solution: Inside a conducting shell the potential is constant and equal to its surface value $\dfrac{kQ}{R}$.
Q12 — Electric Potential & Potential Energy · easy · numerical
The work done in moving a charge of $2\,\mu\text{C}$ through a potential difference of $100\text{ V}$ is:
A. $200\text{ J}$
B. $2 \times 10^{-4}\text{ J}$ ✓ Correct
C. $5 \times 10^{-8}\text{ J}$
D. $2 \times 10^{-6}\text{ J}$
Solution: $W = q\Delta V = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4}\text{ J}$.
Q13 — Electric Potential & Potential Energy · hard · numerical
The electrostatic potential energy of two charges $2\,\mu\text{C}$ and $3\,\mu\text{C}$ placed $0.1\text{ m}$ apart is:
A. $5.4\text{ J}$
B. $54\text{ J}$
C. $0.54\text{ J}$ ✓ Correct
D. $0.054\text{ J}$
Solution: $U = \dfrac{kq_1q_2}{r} = \dfrac{9 \times 10^9 \times 6 \times 10^{-12}}{0.1} = 0.54\text{ J}$.
Q14 — Electric Potential & Potential Energy · hard · numerical
The potential in a region varies as $V = 10x^2\text{ volt}$. The electric field at $x = 2\text{ m}$ is:
A. $-40\text{ V/m}$ ✓ Correct
B. $-10\text{ V/m}$
C. $+40\text{ V/m}$
D. $-20\text{ V/m}$
Solution: $E = -\dfrac{dV}{dx} = -20x$. At $x = 2$, $E = -40\text{ V/m}$.
Q15 — Electric Potential & Potential Energy · medium · numerical
The potential energy of two point charges $+q$ and $-q$ separated by a distance $r$ is:
A. $-\dfrac{kq^2}{r^2}$
B. $-\dfrac{kq^2}{r}$ ✓ Correct
C. $+\dfrac{kq^2}{r}$
D. Zero
Solution: $U = \dfrac{kq_1q_2}{r}$ with $q_1q_2 = -q^2$, so the energy is negative — the pair is bound.
Q16 — Electric Potential & Potential Energy · medium · numerical
An electron is accelerated through a potential difference of $100\text{ V}$. The kinetic energy it gains is:
A. $1.6 \times 10^{-19}\text{ J}$
B. $1.6 \times 10^{-17}\text{ J}$ ✓ Correct
C. $100\text{ J}$
D. $1.6 \times 10^{-21}\text{ J}$
Solution: $KE = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}\text{ J}$, which is $100\text{ eV}$.
Q17 — Electric Potential & Potential Energy · medium · numerical
The potential at $0.5\text{ m}$ from a charge of $5\,\mu\text{C}$ is:
A. $9 \times 10^4\text{ V}$ ✓ Correct
B. $2.25 \times 10^4\text{ V}$
C. $4.5 \times 10^4\text{ V}$
D. $1.8 \times 10^5\text{ V}$
Solution: $V = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{0.5} = \dfrac{4.5 \times 10^4}{0.5} = 9 \times 10^4\text{ V}$.
Q18 — Electric Potential & Potential Energy · medium · numerical
Two equal and opposite charges are separated by a distance $2a$. The potential at the midpoint of the line joining them is:
A. $-\dfrac{kq}{a}$
B. $\dfrac{kq}{a}$
C. $\dfrac{2kq}{a}$
D. Zero ✓ Correct
Solution: The two contributions $+\dfrac{kq}{a}$ and $-\dfrac{kq}{a}$ cancel, since potential is a scalar that adds algebraically.
Q19 — Electric Potential & Potential Energy · medium · numerical
A conducting sphere of radius $0.2\text{ m}$ carries a charge of $2\,\mu\text{C}$. Its surface potential is:
A. $1.8 \times 10^5\text{ V}$
B. $3.6 \times 10^4\text{ V}$
C. $9 \times 10^4\text{ V}$ ✓ Correct
D. $4.5 \times 10^5\text{ V}$
Solution: $V = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.2} = \dfrac{1.8 \times 10^4}{0.2} = 9 \times 10^4\text{ V}$.
Q20 — Electric Potential & Potential Energy · hard · numerical
The work done in bringing a charge of $1\,\mu\text{C}$ from infinity to a point $0.1\text{ m}$ from a charge of $3\,\mu\text{C}$ is:
A. $2.7\text{ J}$
B. $27\text{ J}$
C. $0.27\text{ J}$ ✓ Correct
D. $0.027\text{ J}$
Solution: $W = qV = 10^{-6} \times \dfrac{9 \times 10^9 \times 3 \times 10^{-6}}{0.1} = 10^{-6} \times 2.7 \times 10^5 = 0.27\text{ J}$.
Q21 — Electric Potential & Potential Energy · medium · numerical
If the potential at a point due to a charge is $V$, the potential at twice that distance is:
A. $\dfrac{V}{2}$ ✓ Correct
B. $2V$
C. $4V$
D. $\dfrac{V}{4}$
Solution: $V \propto \dfrac{1}{r}$, so doubling the distance halves the potential (unlike the field, which quarters).