Capacitor — Page 6 | NEET Physics

Q51

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :

A 0.9 n C

B 1.2 n C

C 0.3 n C

D 2.4 n C

Correct Answer

Option B

Solution

Charge on Capacitor initially, Qi = CV After inserting dielectric of dielectric constant = K, new capacitance, Qf = (KC)

\vee

\therefore\,\,\,

Induced charges on dielectric = Qf $-$ Qi = KCV $-$ CV = (K $-$ ) CV =

\left( {{5 \over 3} - 1} \right)

$\times$ 90 $\times$ 10 $-$ 12 $\times$ 20 = 1.2 $\times$ 10 $-$ 9 C = 1.2 nC

Q52

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :

A 4 : 1

B 1 : 2

C 2 : 1

D 1 : 4

Correct Answer

Option D

Solution

Given, C1 = C2 = C When both capacitors are connected in series, their equivalent capacitance will be

{1 \over {{C_s}}} = {1 \over C} + {1 \over C} = {2 \over C}

\Rightarrow {C_s} = {C \over 2}

When both capacitors are connected in parallel, their equivalent capacitance will be Cp = C + C = 2C $\therefore$ The ratio of equivalent capacitance in series and parallel combination is

{{{C_s}} \over {{C_p}}} = {{C/2} \over {2C}} = {1 \over 4}

$\therefore$ Cs : Cp = 1 : 4

Q53

A parallel plate capacitor has plate of length 'l', width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

A

{l \over 4}

B

{l \over 2}

C

{{2l} \over 3}

D

{l \over 3}

Correct Answer

Option D

Solution

Ci =

{{{\varepsilon _0}A} \over d} = {{{\varepsilon _0}lw} \over d}

Ui =

{1 \over 2}{C_i}{V^2}

=

{1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}

Cf = C1 + C2 =

{{K{\varepsilon _0}{A_1}} \over d} + {{{\varepsilon _0}{A_2}} \over d}

=

{{K{\varepsilon _0}wx} \over d} + {{{\varepsilon _0}w\left( {l - x} \right)} \over d}

=

{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]

$\therefore$ Uf =

{1 \over 2}{C_f}{V^2}

=

{1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}

Given Uf = 2Ui $\Rightarrow$

{1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}

= 2 $\times$

{1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}

$\Rightarrow$ kx + l – x = 2l $\Rightarrow$ 4x – x = l $\Rightarrow$ 3x = l $\Rightarrow$ x =

{l \over 3}

Q54

A parallel plate capacitor is formed by two plates each of area 30

\pi

cm2 separated by 1 mm. A material of dielectric strength 3.6

\times

107 Vm

-

1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7

\times

10

-

6C, the value of dielectric constant of the material is : [Use

{1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}

Nm2 C

-

2]

A 1.66

B 1.75

C 2.25

D 2.33

Correct Answer

Option D

Solution

Field inside the dielectric

= {\sigma \over {k{\varepsilon _0}}}

According to the given information,

{\sigma \over {k{\varepsilon _0}}} = 3.6 \times {10^7}

\Rightarrow {{{Q \over A}} \over {k{\varepsilon _0}}} = 3.6 \times {10^7}

\Rightarrow k = 2.33

Q55

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

A Zero

B

{3 \over 2}C{V^2}

C

{9 \over 2}C{V^2}

D

{{25} \over 6}C{V^2}

Correct Answer

Option B

Solution

By conservation of charge,

{q_1} + {q_2} = {q_1}' + {q_2}'

$\Rightarrow$

- CV + (2C)(2V) = (C + 2C)V'

V' = {{3CV} \over {3C}} = V

{U_f} = {1 \over 2}C{v^2} + {1 \over 2}(2C){V^2}

{U_f} = {3 \over 2}C{V^2}

Q56

Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be :

A 4.2 μF

B 8.4 μF

C 1.6 μF

D 3.2 μF

Correct Answer

Option C

Solution

C1 + C2 = 10 ......(1)

{1 \over 2}{C_2}{V^2} = 4 \times {1 \over 2}{C_1}{V^2}

$\Rightarrow$ C2 = 4C1 .....(2) From (1) ans (2), C1 = 2 and C2 = 8 For series combination Ceq =

{{{C_1}{C_2}} \over {{C_1} + {C_2}}}

=

{{8 \times 2} \over {8 + 2}}

= 1.6 $\mu$ F

Q57

A capacitor is connected to a 20 V battery through a resistance of 10

\Omega

. It is found that the potential difference across the capacitor rises to 2 V in 1

\mu

s. The capacitance of the capacitor is __________

\mu

F. Given :

\ln \left( {{{10} \over 9}} \right) = 0.105

A 9.52

B 0.95

C 0.105

D 1.85

Correct Answer

Option B

Solution

Given, the peak voltage of the battery, V0 = 20V The voltage of the battery, V = 2V Time, t = 1 $\mu$ s = 1 $\times$ 10 $-$ 6s Resistance of the capacitor, R = 10

\Omega

As we know that, V = V0(1 $-$ e $-$ t/RC) Substituting the values in the above equation, we get 2 = 20(1 $-$ e $-$ t/RC)

\Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)

\Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95

$\mu$ F $\therefore$ The capacitance of the capacitor is 0.95 $\mu$ F.

Q58

Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is

{{15} \over 4}

times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors,

{{{C_2}} \over {{C_1}}}

.

A

{{15} \over {11}}

B No Solutions

C

{{29} \over {15}}

D

{{15} \over {4}}

Correct Answer

Option B

Solution

When connected in parallel Ceq = C1 + C2 When in series

C{'_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}

{C_1} + {C_2} = {{15} \over 4}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)

4{({C_1} + {C_2})^2} = 15{C_1}{C_2}

4{C_1}^2 + 4{C_2}^2 - 7{C_1}{C_2} = 0

dividing by

{C_1}^2

4{\left( {{{{C_2}} \over {{C_1}}}} \right)^2} - {{7{C_2}} \over {{C_1}}} + 4 = 0

Let

{{{C_2}} \over {{C_1}}} = x

4{x^2} - 7x + 4 = 0

{b^2} - 4ac = 49 - 64 < 0

$\therefore$ No solution exixts.

Q59

A parallel plate capacitor has plate area 40 cm

^2

and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :

A

\mathrm{10\varepsilon_0~F}

B

\mathrm{24\varepsilon_0~F}

C

\mathrm{\dfrac{3}{10}\varepsilon_0~F}

D

\mathrm{\dfrac{10}{3}\varepsilon_0~F}

Correct Answer

Option D

Solution

This can be seen as two capacitors in series combination so

\begin{aligned} & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2} \\\\ & =\frac{1}{\frac{\mathrm{K} \in_0 \mathrm{~A}}{\mathrm{t}}}+\frac{1}{\frac{\in_0 \mathrm{~A}}{\mathrm{~d}-\mathrm{t}}} \end{aligned}

\begin{aligned} & =\frac{\mathrm{t}}{\mathrm{K} \in_0 \mathrm{~A}}+\frac{\mathrm{d}-\mathrm{t}}{\epsilon_0 \mathrm{~A}} \\\\ & =\frac{1 \times 10^{-3}}{5 \in_0 \times 40 \times 10^{-4}}+\frac{1 \times 10^{-3}}{\in_0 40 \times 10^{-4}} \\\\ & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{20 \in_0}+\frac{1}{4 \in_0} \\\\ & \mathrm{C}_{\mathrm{eq}}=\frac{20 \times 4 \in_0}{24}=\frac{10 \in_0}{3} \mathrm{~F} \end{aligned}

Q60

Two capacitors

{C_1}

and

{C_2}

are charged to

120

V

and

200

V

respectively. It is found that connecting them together the potential on each one can be made zero. Then

A

5{C_1} = 3{C_2}

B

3{C_1} = 5{C_2}

C

3{C_1} + 5{C_2} = 0

D

9{C_1} = 4{C_2}

Correct Answer

Option B

Solution

For potential to be made zero, after connection

120{C_1} = 200{C_2}

\left[ {\,\,} \right.

as

\left. {C = {q \over v}\,\,} \right]

\Rightarrow 3{C_1} = 5{C_2}