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

Answer: A

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

Answer: C

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

Answer: B

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

Then, E1 = Q/(ε_{0} × ε_{1} × A)

E2 = Q/(ε_{0} × ε_{2} × A)

where

ε_{0} = 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

Answer: A

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

Answer: C

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/(ε_{0} × ε_{1} × A)

E2 = Q/(ε_{0} × ε_{2} × A)

where

ε_{0} = 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

Answer: B

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

Answer: B

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^{-11}F
b) 7.08 × 10^{11}F
c) 7.08 × 10^{-12}F
d) 7.08 × 10^{-10}F

Answer: A

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

Answer: C

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

Answer: B

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.

Hence as the thickness of the dielectric slab increases, a larger value is subtracted from the original potential difference.