# Capacitance of Multiplate Capacitor MCQ [Free PDF] – Objective Question Answer for Capacitance of Multiplate Capacitor Quiz

11. Potential drop in a dielectric is equal to _______

A. Electric field strength × thickness
B. Electric field strength × area of a cross-section
C. Electric field strength
D. Zero

When a dielectric is introduced between the two plates of a parallel plate capacitor, the potential difference decreases by the value of the product of electric field strength × thickness which is the potential difference of the dielectric.

12. The electric field strength is 10N/C and the thickness of the dielectric is 3m. Calculate the potential drop in the dielectric.

A. 10V
B. 20V
C. 30V
D. 40V

The potential drop in a dielectric = electric field strength × area of cross section

= 10 × 3 = 30V.

13. The electric fields of dielectrics having the same cross-sectional area in series are related to their relative permittivities in which way?

A. Directly proportional
B. Inversely proportional
C. Equal
D. Not related

Let us consider two plates having fields E1 and E2 and relative permittivities e1 and e2.

Then, E1 = Q/(ε0 × ε1 × A)

E2 = Q/(ε0 × ε2 × A)

where

ε0 = absolute permittivity

A = area of cross section.

From the given expression, we see that E1/E2 = ε2/ε1

Hence the electric field is inversely proportional to the relative permittivities.

14. What happens to the capacitance when a dielectric is introduced between its plates?

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero

The capacitance of a capacitance increases when a dielectric is introduced between its plates because the capacitance is related to the dielectric constant k by the equation:

C = k∈0A/d.

15. Calculate the relative permittivity of the second dielectric if the relative permittivity of the first is 4. The electric field strength of the first dielectric is 8V/m and that of the second is 2V/m.

A. 32
B. 4
C. 16
D. 8

The relation between the two electric fields and the relative permittivities is:

Let us consider two plates having fields E1 and E2 and relative permittivities e1 and e2.

Then, E1 = Q/(ε0 × ε1 × A)

E2 = Q/(ε0 × ε2 × A)

where

ε0 = absolute permittivity

A = area of cross section.

From the given expression, we see that E1/E2 = ε2/ε1

Substituting the given values, we get e2 = 16.

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16. What happens to the potential drop between the two plates of a capacitor when a dielectric is introduced between the plates?

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero

When a dielectric is introduced between the two plates of a parallel plate capacitor, the potential difference decreases because the potential difference of the dielectric is subtracted from it.

17. If the potential difference across the plates of a capacitor is 10V and a dielectric having a thickness of 2m is introduced between the plates, calculate the potential difference after introducing the dielectric. The electric field strength is 2V/m.

A. 4V
B. 6V
C. 8V
D. 10V

When a dielectric is introduced between the plates of a capacitor, its potential difference decreases.

New potential difference = potential difference without dielectric − potential difference of dielectric

= 10-2 × 2 = 6V.

18. Calculate the capacitance if the dielectric constant = 4, area of cross-section = 10m2, and the distance of separation between the plates are 5m.

a) 7.08 × 10-11F
b) 7.08 × 1011F
c) 7.08 × 10-12F
d) 7.08 × 10-10F

The expression to find capacitance when a dielectric is introduced between the plates is:

C = ke0A/d.

Substituting the given values in the equation, we get

C = 7.08 × 10-11F.

19. A dielectric is basically a ________

A. Capacitor
B. Conductor
C. Insulator
D. Semiconductor

A dielectric is an insulator because it has all the properties of an insulator.

20. What happens to the potential difference between the plates of a capacitor as the thickness of the dielectric slab increases?

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero