# Physics Questions and Answers – Parallel Plate Capacitor

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Parallel Plate Capacitor”.

1. Identify the simplest and the most widely used capacitor among the following.
a) Electrolytic capacitor
b) Spherical Capacitor
c) Parallel plate capacitor
d) Cylindrical capacitor

Explanation: The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a small distance.

2. How is the electric field between the two plates of a parallel plate capacitor?
a) Zero
b) Uniform
c) Maximum
d) Minimum

Explanation: The direction of the electric field is from the positive to the negative plate. In the inner region, between the two capacitor plates, the electric fields due to the two charged plates add up. Hence, the field is uniform throughout.

3. Identify the factor on which the capacitance of a parallel plate capacitor does not depend.
a) Permeability of the medium between the plates
b) Area of the plates
c) Distance between the plates
d) The permittivity of the medium between the plates

Explanation: The capacitance of a parallel plate capacitor is directly proportional to the area of the plates and permittivity of the medium between the plates. It is indirectly proportional to the distance between the plates.
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4. Calculate the capacitance of the capacitor, if 1012 electrons are transferred from one conductor to another of a capacitor and a potential difference of 10 V develops between the two conductors.
a) 1.6 × 10-7 F
b) 160 × 10-8 F
c) 16 × 10-8 F
d) 1.6 × 10-8 F

Explanation: q = ne = 1012 × 1.6 × 10-19 = 1.6 × 10-7 C.
V = 10 V.
C = $$\frac {q}{V}$$ = 1.6 × $$\frac {1.6}{10^{-7}}$$

C = 1.6 × 10-8 F.

5. What is the net electric field in the outer regions above the upper plate and below the lower plate in a parallel plate capacitor?
a) Maximum
b) Uniform
c) Zero
d) Minimum

Explanation: In the outer regions above the upper plate and below the lower plate, the electric fields due to the two charged plates cancel out. Hence, the net electric field in the outer regions above the plate and below the lower plate is zero.

6. In the inner region between the two capacitor plates, the electric fields due to the two charged plates are zero.
a) True
b) False

Explanation: In the inner region between the two capacitor plates, the electric fields due to the two charged plates add up. The net field is given by:
$$\frac {\sigma}{2\varepsilon_0}+\frac {\sigma}{2\varepsilon_0}=\frac {\sigma}{\varepsilon_0}$$.

7. ‘X’ is a widely used capacitor which consists of two large plane parallel conducting plates separated by a small distance. Identify X.
a) Spherical capacitor
b) Parallel plate capacitor
c) Cylindrical capacitor
d) Electrolytic capacitor

Explanation: The simplest and the most widely used capacitor is the parallel plate capacitor. It consists of two large plane parallel conducting plates, separated by a small distance.
Capacitance of a parallel plate capacitor = $$\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}$$.

8. Find out the correct expression of the capacitance of a parallel plate capacitor where ‘A’ is the area of the plates, ‘d’ is the distance between the plates and ‘ε0’is the permittivity of the medium.
a) $$\frac {A\varepsilon_0}{d}$$
b) $$\frac {Ad}{\varepsilon_0}$$
c) $$\frac {d\varepsilon_0}{A}$$
d) Aε0d

Explanation: The capacitance of a parallel plate capacitor is directly proportional to the area of the plates and permittivity of the medium between the plates. It is indirectly proportional to the distance between the plates.
C = $$\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}$$.

9. A parallel plate capacitor has a plate area of 100 cm2 and is separated by a distance of 20 mm. Find its capacitance.
a) 6.425 × 10-12 F
b) 5.425 × 10-12 F
c) 4.425 × 10-12 F
d) 3.425 × 10-12 F

Explanation: C = $$\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}$$.
C = $$\frac {100 \times 10^{-4} \times 8.85 \times 10^{-12})}{(20 \times 10^{-3})}$$
C = 4.425 × 10-12 F.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

To practice all areas of Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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