Capacitor — Page 2 | JEE Physics

Q11

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity

10^6 \mathrm{~m} / \mathrm{s}

. If the magnitude of the electric field between the plates is

9.1 \mathrm{~V} / \mathrm{cm}

, then the vertical component of velocity of electron is (mass of electron

=9.1 \times 10^{-31} \mathrm{~kg}

and charge of electron

=1.6 \times 10^{-19} \mathrm{C}

)

A

1 \times 10^6 \mathrm{~m} / \mathrm{s}

B

16 \times 10^6 \mathrm{~m} / \mathrm{s}

C

16 \times 10^4 \mathrm{~m} / \mathrm{s}

D 0

Correct Answer

Option B

Solution

No force in horizontal direction, i.e., $F_x=0$ $\Rightarrow a_x=0$ So, $v_x$ = constant hence, $t=\dfrac{l}{v_x}$ .....

(1) Now, for y-direction, using Ist equation of motion,

{v_y} = {u_y} + {a_y}t

\Rightarrow {v_y} = 0 + {{eE} \over m}\left( {{l \over {{v_x}}}} \right)

(from (1) and usiung $F=ma$ and $a=\dfrac{F}{m}$ )

\Rightarrow {v_y} = {{1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^2} \times 0.1} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}

= 1.6 \times {10^7}

m/s

\Rightarrow {v_y} = 16 \times {10^6}

m/s.

Q12

Three capacitors each of 4

\mu

F are to be connected in such a way that the effective capacitance is 6

\mu

F. This can be done by connecting them :

A all in series

B two in series and one in parallel

C all in parallel

D two in parallel and one in series

Correct Answer

Option B

Solution

(a)

{1 \over {{C_{eq}}}} = {1 \over 4} + {1 \over 4} + {1 \over 4} = {3 \over 4}

$\Rightarrow$

{C_{eq}} = {4 \over 3}\,\mu F

(b)

{C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F

(c)

{C_{eq}} = 4 + 4 + 4 = 12\,\mu F

(d)

{C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F

Q13

The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4

\mu

C charge, its radius will be : [ Take :

{1 \over {4\,\pi { \in _0}}} =

9

\times

109 N

-

m2/C2 ]

A 20 mm

B 32 mm

C 28 mm

D 16 mm

Correct Answer

Option D

Solution

Energy of sphere =

{{{Q^2}} \over {2C}}

\therefore\,\,\,

{{16 \times {{10}^{ - 12}}} \over {2C}}

= 4.5 $\Rightarrow$

\,\,\,

C =

{{16 \times {{10}^{ - 12}}} \over 9}

We know capacity of spherical conductor, C = 4 $\pi$

\varepsilon

0R

\therefore\,\,\,

4 $\pi$

\varepsilon

0R =

{{16 \times {{10}^{ - 12}}} \over 9}

$\Rightarrow$

\,\,\,

R =

{1 \over {4\pi {\varepsilon _0}}} \times {{16 \times {{10}^{ - 12}}} \over 9}

= 9 $\times$ 109 $\times$

{{16 \times {{10}^{ - 12}}} \over 9}

= 16 mm

Q14

A parallel plate capacitor of capacitance

2 \mathrm{~F}

is charged to a potential

\mathrm{V}

, The energy stored in the capacitor is

E_{1}

. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is

\mathrm{E}_{2}

. The ratio

\mathrm{E}_{2} / \mathrm{E}_{1}

is :

A 1 : 2

B 2 : 3

C 2 : 1

D 1 : 4

Correct Answer

Option A

Solution

To determine the ratio

\mathrm{E}_{2} / \mathrm{E}_{1}

, we will follow these steps: 1. Calculate the initial energy stored in the original capacitor

\mathrm{E}_{1}

. 2. Determine the energy stored in the system when two capacitors are connected in parallel, which is

\mathrm{E}_{2}

. 3. Find the ratio

\mathrm{E}_{2} / \mathrm{E}_{1}

. Step 1: Calculate the initial energy stored in the original capacitor

\mathrm{E}_{1}

. The energy stored in a capacitor is given by the formula:

E = \frac{1}{2} CV^2

Given that the capacitance

C

is

2 \mathrm{~F}

, and the potential difference is

\mathrm{V}

, the initial energy

\mathrm{E}_{1}

is:

E_{1} = \frac{1}{2} \cdot 2 \mathrm{~F} \cdot V^2

E_{1} = V^2 \mathrm{~J}

Step 2: Determine the energy stored when two capacitors are connected in parallel When the charged capacitor (capacitor 1) is connected to an identical uncharged capacitor (capacitor 2), the charge will redistribute between the two capacitors.

The total capacitance of the parallel combination is:

C_{\text{total}} = 2 \mathrm{~F} + 2 \mathrm{~F} = 4 \mathrm{~F}

The initial charge on capacitor 1 is:

Q_1 = CV = 2 \mathrm{~F} \cdot V = 2V \mathrm{~C}

After connection, this charge will be shared equally by the two capacitors because they are identical.

Therefore, the voltage across each capacitor in the parallel combination will be:

V_{\text{across each capacitor}} = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{2V}{4 \mathrm{~F}} = \frac{V}{2}

The energy stored in the parallel combination is:

E_{2} = \frac{1}{2} \cdot 4 \mathrm{~F} \cdot \left(\frac{V}{2}\right)^2

E_{2} = \frac{1}{2} \cdot 4 \mathrm{~F} \cdot \frac{V^2}{4}

E_{2} = V^2 \cdot \frac{1}{2} \mathrm{~J}

Step 3: Find the ratio

\mathrm{E}_{2} / \mathrm{E}_{1}

We know that:

E_{1} = V^2 \mathrm{~J}

E_{2} = \frac{1}{2} V^2 \mathrm{~J}

The ratio is then:

\frac{E_{2}}{E_{1}} = \frac{\frac{1}{2} V^2}{V^2} = \frac{1}{2} = 1 : 2

Therefore, the correct answer is: Option A: 1 : 2

Q15

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

Q16

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

Q17

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}

Q18

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

Q19

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}

Q20

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