200+ Capacitance and Capacitor MCQ – Objective Question Answer for Capacitance and Capacitor Quiz

101. Which is the correct expression for the capacitance of a multi-plate capacitor?

A. C = absolute permittivity × A/d
B. C = Actual permittivity × (n-1) × A/d
C. C = Actual permittivity × (n) × A/d
D. Actual permittivity × (n+1) × A/d

Answer: B

The correct expression for capacitance of a multi plate capacitor is

C = Actual permittivity × (n-1) × A/d.

Where

n = number of plates
A = area of cross section of the plates
d = distance of separation between the plates.

 

102. What happens to the capacitance of a multi-plate capacitor when the area of a cross-section of the plate decreases?

A. Increases
B. Decreases
C. Remains constant
D. Becomes zero

Answer: B

When the area of the cross-section decreases, the capacitance also decreases since it is related to the formula

C = Actual permittivity × (n-1) × A/d.

Here, we can see that the capacitance is directly proportional to the area of the cross-section.

 

103. What happens to the capacitance of a multi-plate capacitor when the distance of separation between the plate increases?

A. Increases
B. Decreases
C. Remains constant
D. Becomes zero

Answer: B

When the distance of separation between the plates increases, the capacitance decreases since it is related to the formula

C = Actual permittivity × (n-1) × A/d.

Here, we can see that the capacitance is inversely proportional to the distance of separation.

 

104. What happens to the capacitance of a multi-plate capacitor when the number of plates increases?

A. Increases
B. Decreases
C. Remains constant
D. Becomes zero

Answer: A

When the number of capacitors increases, the capacitance also increases since it is related to the formula

C = Actual permittivity × (n-1) × A/d.

Here, we can see that the capacitance is directly proportional to the number of capacitors.

 

105. Find the capacitance of a multi plate capacitor whose actual permittivity = 5F/m, n = 3, A = 4m2and d = 2m.

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

Answer: B

The formula for capacitance of a multi plate capacitor

C = Actual permittivity × (n-1) × A/d.

Thus, C = 5 × (3-1) × 4/2 = 20F.

 

106. Find the capacitance of a multi plate capacitor whose relative permittivity = 5, n = 3, A = 4m2 and d = 2m.

A. 1.77 × 10-10 F
B. 1.77 × 1010 F
C. 1.77 × 10-11 F
D. 1.77 × 1011 F

Answer: A

The formula for capacitance of a multi plate capacitor

C = Relative permittivity × absolute permittivity × (n-1) × A/d.

C = 5 × 8.85 × 10-12 × (3-1) × 4/2 = 1.77 × 10-10.

 

107. Find the number of plates in the multi plate capacitor having C = 20F absolute permittivity = 5F/m, A = 4m2 and d = 2m.

A. 1
B. 2
C. 3
D. 4

Answer: C

The formula for capacitance of a multi plate capacitor:

C = Actual permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get n = 3.

 

108. Calculate the distance between the plates of the capacitor having C = 20F, actual permittivity = 5F/m n = 3 and A = 4m2.

A. 1m
B. 2m
C. 3m
D. 4m

Answer: B

The formula for the capacitance of a multi-plate capacitor

C = Actual permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get d = 2m.

 

109. Calculate the area of cross section of the multi plate capacitor having C = 20F, actual permittivity = 5F/m n = 3 and d = 2m.

A. 1m2
B. 2m2
C. 3m2
D. 4m2

Answer: D

The formula for the capacitance of a multi-plate capacitor

C = Actual permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get A = 4m2.

 

110. Calculate the number of plates in the multi plate capacitor having C = 1.77 × 10-10F relative permittivity = 5, A = 4m2 and d = 2m.

A. 1
B. 2
C. 3
D. 4

Answer: C

The formula for the capacitance of a multi-plate capacitor

C = Relative permittivity × absolute permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get n = 3.

 

111. 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.

 

112. 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.

 

113. 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/(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.

 

114. 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.

 

115. 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/(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.

 

116. 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.

 

117. 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.

 

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

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

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.

 

119. 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.

 

120. 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.

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