Practice Start Test

Capacitor

NEET Physics · 98 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to
A C4{C \over 4}
B 2C
C C2{C \over 2}
D 4C
Correct Answer
Option A
Solution

For a capacitor of area A and distance between the plates as d, the capacitance (C) is

C=A∈0dC = {{A{ \in _0}} \over d}

On doubling the distance and reducing area to half

C1=A2∈02d{C_1} = {{{A \over 2}{ \in _0}} \over {2d}}
⇒C1=A∈04d\Rightarrow {C_1} = {{A{ \in _0}} \over {4d}}
⇒C1=C4\Rightarrow {C_1} = {C \over 4}
Q12
The effective capacitances of two capacitors are 3 μ\muF and 16 μ\muF, when they are connected in series and parallel respectively. The capacitance of two capacitors are :
A 1.2 μ\muF, 1.8 μ\muF
B 10 μ\muF, 6 μ\muF
C 8 μ\muF, 8 μ\muF
D 12 μ\muF, 4 μ\muF
Correct Answer
Option D
Solution

Here,

1C1+1C2=13{1 \over {{C_1}}} + {1 \over {{C_2}}} = {1 \over 3}

(Series combination) While,

C1+C2=16{C_1} + {C_2} = 16

(Parallel combination) Solving the above equations we get

C1=12 μF{C_1} = 12\,\mu F

and

C2=4 μF{C_2} = 4\,\mu F
Q13
A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is
A 4.5 ×\times 10 −-6 J
B 3.25 ×\times 10 −-6 J
C 2.25 ×\times 10 −-6 J
D 1.5 ×\times 10 −-6 J
Correct Answer
Option C
Solution

q 1 = CV = 900 ×\times 10 −-12 ×\times 100 = 9 ×\times 10 −-8 C

V=C1V1+C2V2C1+C2V = {{{C_1}{V_1} + {C_2}{V_2}} \over {{C_1} + {C_2}}}
=9×10−8+01800×10−12=1002=50= {{9 \times {{10}^{ - 8}} + 0} \over {1800 \times {{10}^{ - 12}}}} = {{100} \over 2} = 50

V

U=12(C1+C2)V2U = {1 \over 2}({C_1} + {C_2}){V^2}
=12×1800×10−12×50×50= {1 \over 2} \times 1800 \times {10^{ - 12}} \times 50 \times 50
=225×10−8= 225 \times {10^{ - 8}}
U=2.25×10−6U = 2.25 \times {10^{ - 6}}

J

Q14
A parallel plate capacitor has a uniform electric field E→\overrightarrow E in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is : (ε\varepsilon 0 = permittivity of free space)
A E2Adε0{{{E^2}Ad} \over {{\varepsilon _0}}}
B 12ε0E2{1 \over 2}{\varepsilon _0}{E^2}
C ε0EAd{\varepsilon _0}EAd
D 12ε0E2Ad{1 \over 2}{\varepsilon _0}{E^2}Ad
Correct Answer
Option D
Solution

Energy = Energy density ×\times volume =

12ε0E2Ad{1 \over 2}{\varepsilon _0}{E^2}Ad
Q15
The equivalent capacitance of the combination shown in the figure is :
A 3C/2
B 3C
C 2C
D C/2
Correct Answer
Option C
Solution

C AB = 2C

Q16
The capacitance of a parallel plate capacitor with air as medium is 6μ\mu F. With the introduction of a dielectric medium, the capacitance become 30 μ\mu F The permittivity of the medium is : (ε0=8.85×10−12C2N−1m−2)\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}} \right)
A 1.77×10−12C2N−1m−21.77 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}
B 0.44×10−10C2N−1m−20.44 \times {10^{ - 10}}{C^2}{N^{ - 1}}{m^{ - 2}}
C 5.00C2N−1m−25.00{C^2}{N^{ - 1}}{m^{ - 2}}
D 0.44×10−13C2N−1m−20.44 \times {10^{ - 13}}{C^2}{N^{ - 1}}{m^{ - 2}}
Correct Answer
Option B
Solution

The capacitance increases with dielectric constant K

C=KC0C = K{C_0}

(or)

K=CC0=306=5K = {C \over {{C_0}}} = {{30} \over 6} = 5
ε=Kε0=\varepsilon = K{\varepsilon _0} =
5×8.85×10−12=5 \times 8.85 \times {10^{ - 12}} =
0.44×10−100.44 \times {10^{ - 10}}
Q17
A parallel plate capacitor of capacitance 20 μ\mu F is being charged by a voltage source whose potential is changing at the rate of 3V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively :
A zero, zero
B zero, 60 μ\mu A
C 60 μ\mu A, 60 μ\mu A
D 60 μ\mu A, zero
Correct Answer
Option C
Solution

As, q = CV Differentiating above equation w.r.t. t, we get

dqdt=CdVdt{{dq} \over {dt}} = C{{dV} \over {dt}}

⇒\Rightarrow i c =

CdVdtC{{dV} \over {dt}}

[As i =

dqdt{{dq} \over {dt}}

] = 20 ×\times 10 -6 ×\times 3 = 60 ×\times 10 -6 A = 60μ\muA Also, conduction current in wires is equal to displacement current between the plates of capacitor. ∴\therefore i d = i c = 60 μ\muA

Q18
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
A Independent of the distance between the plates
B Linearly proportional to the distance between the plates
C Proportional to the square root of the distance between the plates
D Inversely proportional to the distance between the plates
Correct Answer
Option A
Solution

In an isolated capacitor Q is constant, so an electrostatic force between metal plates is given as: F = QE = Q ×\times

σ2ε0{\sigma \over {2{\varepsilon _0}}}

=

Q22Aε0{{{Q^2}} \over {2A{\varepsilon _0}}}

[∵\because

σ=QA\sigma = {Q \over {A}}

] F is independent of the distance between plates.

Q19
A capacitor is charged by a battery. The battery is removed and another identical unchanged capacitor is connected in parallel. The total electrostatic energy of resulting system
A decreases by a factor of 2
B remains the same
C increases by a factor of 2
D increases by a factor of 4
Correct Answer
Option A
Solution

When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,

Ui=12CV2{U_i} = {1 \over 2}C{V^2}

...(i) When the battery is removed and another identical uncharged capacitor is connected in parallel Common potential,

V′=CVC+C=V2V' = {{CV} \over {C + C}} = {V \over 2}

Then the energy stored in the capacitor,

Uf=12(2C)(V2)2=14CV2{U_f} = {1 \over 2}\left( {2C} \right){\left( {{V \over 2}} \right)^2} = {1 \over 4}C{V^2}

...(ii) ∴\therefore From eqns. (i) and (ii)

Uf=Ui2{U_f} = {{{U_i}} \over 2}

that means the total electrostatic energy of resulting system will decreases by a factor of 2.

Q20
A parallel-plate capacitor of area a, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k 1 , k 2 , k 3 and k 4 as shown in the figure. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by
A k = k 1 + k 2 + k 3 + 3k 4
B k = 23{2 \over 3} (k 1 + k 2 + k 3 ) + 2k 4
C 2k{2 \over k} = 3k1+k2+k3+1k4{3 \over {{k_1} + {k_2} + {k_3}}} + {1 \over {{k_4}}}
D 1k{1 \over k} = 1k1+1k2+1k3+32k4{1 \over {{k_1}}} + {1 \over {{k_2}}} + {1 \over {{k_3}}} + {3 \over {2{k_4}}}
Correct Answer
Option C
Solution

Here,

C1=2ε0k1A3d,C2=2ε0k2A3d{C_1} = {{2{\varepsilon _0}{k_1}A} \over {3d}},{C_2} = {{2{\varepsilon _0}{k_2}A} \over {3d}}
C3=2ε0k3A3d,C4=2ε0k4Ad{C_3} = {{2{\varepsilon _0}{k_3}A} \over {3d}},{C_4} = {{2{\varepsilon _0}{k_4}A} \over d}

Given system of C 1 , C 2 , C 3 and C 4 can be simplified as ∴\therefore

1CAB=1C1+C2+C3+1C4{1 \over {{C_{AB}}}} = {1 \over {{C_1} + {C_2} + {C_3}}} + {1 \over {{C_4}}}

Suppose,

CAB=kε0Ad{C_{AB}} = {{k{\varepsilon _0}A} \over d}
CAB=1k(ε0Ad)=123ε0Ad(k1+k2+k3)+12ε0Adk4{C_{AB}} = {1 \over {k\left( {{{{\varepsilon _0}A} \over d}} \right)}} = {1 \over {{2 \over 3}{{{\varepsilon _0}A} \over d}\left( {{k_1} + {k_2} + {k_3}} \right)}} + {1 \over {{{2{\varepsilon _0}A} \over d}{k_4}}}
⇒1k=12(k1+k2+k3)+12k4\Rightarrow {1 \over k} = {1 \over {2\left( {{k_1} + {k_2} + {k_3}} \right)}} + {1 \over {2{k_4}}}
∴2k=3k1+k2+k3+1k4\therefore {2 \over k} = {3 \over {{k_1} + {k_2} + {k_3}}} + {1 \over {{k_4}}}
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