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
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.