Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 9

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Weekly Quiz Competition

Question 1
4 / -1

If dielectric constant and dielectric strength be denoted by K and X respectively, then a material suitable for the use as a dielectric in a capacitor must have

A
high K and high X

B
high K and low X

C
low K and high X

D
low K and low X

Solution

The material suitable for using as a dielectric must have high dielectric strength X and large dielectric constant K.
Question 2
4 / -1

A parallel plate capacitor with air between the plates has a capacitance of 10 pF. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 4 is

A
80pF

B
96pF

C
100pF

D
120pF

Solution

Here C₁= ε₀A/d = 10pF …….(i)
(∵ K = 4)
= 8 × 10 (using (i))
= 80 pF
Question 3
4 / -1

The capacitance of a parallel plate capacitor with air as medium is 3μF. With the introduction of a dielectric medium between the plates, the capacitance becomes 15μF. The permittivity of the medium is

A
5C²N⁻¹m⁻²

B
10C²N⁻¹m⁻²

C
0.44 × 10⁻¹⁰C²N⁻¹m⁻²

D
8.854 × 10⁻¹¹C²N⁻¹m⁻²

Solution

Capacitance of a parallel capacitor with air is C = ε₀A/d
Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is C ′ = Kε₀A/d
where K is the dielectric constant of a medium
or C′/C = K or K = 15/3 = 5 or K = ε₀/ε
or ε = Kε₀ = 5 × 8.854 × 10⁻¹² = 0.44 × 10⁻¹⁰C²N⁻¹ m⁻²
Question 4
4 / -1

A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is

A

B

C

D

Solution

As capacitance C₀ = ε₀A/d
∴ after inserting copper plate C = ε₀A/d−b
Question 5
4 / -1

A parallel plate capacitor of capacitance 5μF and plate separation 6 cm is connected to a 1 v battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

A
2μc

B
3μc

C
5μc

D
10μc

Solution

as Q = cv
initially c = 5μF, v = 1v
∴ Q = 5 × 10^–6 c = 5μc
when dielectric is introduced, let capacitance be c'
∴ c' = [(A∈₀)/{d – t + (t/k)}]
here k = 4, t = 4 cm, d = 6 cm
∴ c' = [{(A∈₀)/d}/{{d – t + (t/k)}/d}] ------ dividing by d
As [(A∈₀)/d] = initially capacitance = c = 5μF
∴ c' = [c/{{6 – 4 + (4/4)}/6}] = [c/(3/6)] = 2c = 10μF
∴ Q' = c'v = 10 × 10^–6 × 1 = 10μc
∴ addition charge flowing = Q' – Q = 10 – 5 = 5μc