Capacitors & Capacitance — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Capacitors & Capacitance 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 — Capacitors & Capacitance · easy · theory
The capacitance of a conductor is defined as:
A. The work done per unit charge
B. The charge stored per unit area
C. The potential per unit charge
D. The charge required to raise its potential by one volt ✓ Correct
Solution: $C = \dfrac{Q}{V}$, measured in farad. One farad is an enormous capacitance, so practical units are $\mu\text{F}$ and $\text{pF}$.
Q2 — Capacitors & Capacitance · easy · theory
The capacitance of a parallel plate capacitor with plate area $A$ and separation $d$ in vacuum is:
A. $\dfrac{\varepsilon_0A}{d}$ ✓ Correct
B. $\dfrac{\varepsilon_0d}{A}$
C. $\varepsilon_0Ad$
D. $\dfrac{A}{\varepsilon_0d}$
Solution: Bringing the plates closer or enlarging them increases the charge the capacitor can hold at a given voltage.
Q3 — Capacitors & Capacitance · easy · theory
For capacitors connected in series, the equivalent capacitance satisfies:
A. $C = C_1 + C_2 + \ldots$
B. $C = \sqrt{C_1C_2}$
C. $C = C_1C_2$
D. $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots$ ✓ Correct
Solution: In series each capacitor carries the same charge while the potential differences add, so the equivalent capacitance is less than the smallest one.
Q4 — Capacitors & Capacitance · easy · theory
For capacitors connected in parallel, the equivalent capacitance is:
A. $\dfrac{C_1C_2}{C_1 + C_2}$
B. $C_1 + C_2 + \ldots$ ✓ Correct
C. $\dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots$
D. $\sqrt{C_1C_2}$
Solution: All capacitors share the same potential difference while the charges add, so the capacitances simply sum.
Q5 — Capacitors & Capacitance · easy · theory
Introducing a dielectric of constant $K$ between the plates of a capacitor changes its capacitance to:
A. $C$
B. $K^2C$
C. $KC$ ✓ Correct
D. $\dfrac{C}{K}$
Solution: Polarisation of the dielectric reduces the effective field, allowing more charge to be stored at the same voltage.
Q6 — Capacitors & Capacitance · hard · theory
A parallel plate capacitor is charged with the battery still connected, and a dielectric slab is then inserted. The electric field between the plates:
A. Becomes zero
B. Increases by a factor $K$
C. Remains unchanged, since $V$ and $d$ are both fixed ✓ Correct
D. Decreases by a factor $K$
Solution: With the battery connected the potential difference is held constant, and since $E = \dfrac{V}{d}$ with $d$ unchanged, the field is unchanged. The charge increases instead.
Q7 — Capacitors & Capacitance · medium · theory
A charged capacitor is disconnected from the battery. The quantity that then remains constant is the:
A. Capacitance
B. Charge on the plates ✓ Correct
C. Electric field energy
D. Potential difference
Solution: With nowhere for the charge to flow, $Q$ is fixed; changing the geometry then alters $V$ and the stored energy.
Q8 — Capacitors & Capacitance · easy · theory
The capacitance of a parallel plate capacitor increases when:
A. The charge on the plates is increased
B. The separation is increased
C. The plate area is increased or the separation decreased ✓ Correct
D. The plate area is decreased
Solution: From $C = \dfrac{\varepsilon_0A}{d}$, capacitance depends only on geometry and the medium, never on the charge stored.
Q9 — Capacitors & Capacitance · easy · theory
The farad is:
A. A unit of charge
B. A unit of potential
C. A very large unit, so practical capacitors are rated in microfarad or picofarad ✓ Correct
D. A very small unit
Solution: A one-farad capacitor would hold one coulomb at one volt, which is impractically large for ordinary components.
Q10 — Capacitors & Capacitance · medium · numerical
Three capacitors of $2\,\mu\text{F}$, $3\,\mu\text{F}$ and $6\,\mu\text{F}$ are joined in series. The equivalent capacitance is:
A. $1\,\mu\text{F}$ ✓ Correct
B. $11\,\mu\text{F}$
C. $0.5\,\mu\text{F}$
D. $2\,\mu\text{F}$
Solution: $\dfrac{1}{C} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1$, so $C = 1\,\mu\text{F}$.
Q11 — Capacitors & Capacitance · easy · numerical
Three capacitors of $2\,\mu\text{F}$, $3\,\mu\text{F}$ and $6\,\mu\text{F}$ are joined in parallel. The equivalent capacitance is:
A. $6\,\mu\text{F}$
B. $11\,\mu\text{F}$ ✓ Correct
C. $36\,\mu\text{F}$
D. $1\,\mu\text{F}$
Solution: In parallel the capacitances add: $2 + 3 + 6 = 11\,\mu\text{F}$.
Q12 — Capacitors & Capacitance · easy · numerical
A parallel plate air capacitor of $8\,\mu\text{F}$ is completely filled with mica of dielectric constant $6$. Its capacitance becomes:
A. $24\,\mu\text{F}$
B. $1.33\,\mu\text{F}$
C. $48\,\mu\text{F}$ ✓ Correct
D. $14\,\mu\text{F}$
Solution: $C' = KC = 6 \times 8 = 48\,\mu\text{F}$.
Q13 — Capacitors & Capacitance · easy · numerical
Two identical capacitors each of capacitance $C$ are connected in series. The equivalent capacitance is:
A. $C$
B. $4C$
C. $\dfrac{C}{2}$ ✓ Correct
D. $2C$
Solution: $\dfrac{1}{C_{eq}} = \dfrac{1}{C} + \dfrac{1}{C} = \dfrac{2}{C}$, so $C_{eq} = \dfrac{C}{2}$.
Q14 — Capacitors & Capacitance · hard · numerical
A parallel plate capacitor has plate area $0.02\text{ m}^2$ and separation $1\text{ mm}$ in air. Its capacitance is:
A. $1.77 \times 10^{-10}\text{ F}$ ✓ Correct
B. $8.85 \times 10^{-12}\text{ F}$
C. $1.77 \times 10^{-13}\text{ F}$
D. $2.26 \times 10^{9}\text{ F}$
Solution: $C = \dfrac{\varepsilon_0A}{d} = \dfrac{8.85 \times 10^{-12} \times 0.02}{10^{-3}} = 1.77 \times 10^{-10}\text{ F}$.
Q15 — Capacitors & Capacitance · easy · numerical
A capacitor of $20\,\mu\text{F}$ is charged to $500\text{ V}$. The charge stored is:
A. $0.01\text{ C}$ ✓ Correct
B. $0.1\text{ C}$
C. $25\text{ C}$
D. $10^{-4}\text{ C}$
Solution: $Q = CV = 20 \times 10^{-6} \times 500 = 10^{-2}\text{ C}$.
Q16 — Capacitors & Capacitance · easy · numerical
An isolated charged parallel plate capacitor has its plate separation doubled. Its capacitance becomes:
A. Four times as large
B. Half as large ✓ Correct
C. Unchanged
D. Twice as large
Solution: $C = \dfrac{\varepsilon_0A}{d} \propto \dfrac{1}{d}$, so doubling $d$ halves the capacitance.
Q17 — Capacitors & Capacitance · hard · numerical
A dielectric slab of constant $K = 4$ and thickness $\dfrac{d}{2}$ is inserted into a parallel plate capacitor of capacitance $C$ and separation $d$. The new capacitance is:
A. $\dfrac{5}{8}C$
B. $\dfrac{4}{3}C$
C. $2C$
D. $\dfrac{8}{5}C$ ✓ Correct
Solution: $C' = \dfrac{\varepsilon_0A}{d - t + \frac{t}{K}} = \dfrac{\varepsilon_0A}{\frac{d}{2} + \frac{d}{8}} = \dfrac{\varepsilon_0A}{\frac{5d}{8}} = \dfrac{8}{5}C$.
Q18 — Capacitors & Capacitance · hard · numerical
Two identical capacitors each of capacitance $C$ are joined in series, and one of them is completely filled with a dielectric of constant $3$. The equivalent capacitance is:
A. $0.75C$ ✓ Correct
B. $1.5C$
C. $3C$
D. $4C$
Solution: The filled capacitor becomes $3C$, so $C_{eq} = \dfrac{3C \times C}{3C + C} = \dfrac{3C}{4} = 0.75C$.
Q19 — Capacitors & Capacitance · hard · numerical
A $5\,\mu\text{F}$ capacitor charged to $100\text{ V}$ is connected across an uncharged $5\,\mu\text{F}$ capacitor. The common potential is:
A. $200\text{ V}$
B. $50\text{ V}$ ✓ Correct
C. $100\text{ V}$
D. $25\text{ V}$
Solution: Charge is conserved: $V = \dfrac{Q_{total}}{C_{total}} = \dfrac{5 \times 10^{-6} \times 100}{10 \times 10^{-6}} = 50\text{ V}$.
Q20 — Capacitors & Capacitance · hard · numerical
Two identical capacitors are joined in parallel and charged to potential $V$. After disconnecting the battery, a dielectric of constant $3$ is inserted into one of them. The new common potential is:
A. $\dfrac{V}{3}$
B. $\dfrac{2V}{3}$
C. $\dfrac{V}{2}$ ✓ Correct
D. $\dfrac{V}{4}$
Solution: Total charge $2CV$ is conserved, while the capacitance becomes $3C + C = 4C$. Hence $V' = \dfrac{2CV}{4C} = \dfrac{V}{2}$.
Q21 — Capacitors & Capacitance · easy · numerical
Two identical capacitors each of capacitance $C$ are connected in parallel. The equivalent capacitance is:
A. $\dfrac{C}{2}$
B. $4C$
C. $C$
D. $2C$ ✓ Correct
Solution: Capacitances in parallel add: $C + C = 2C$.