Capacitor — Page 5 | NEET Physics

Q41

The total charge on the system of capacitors

C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}

and

\mathrm{C}_{4}=3 \mu \mathrm{F}

connected in parallel is : (Assume a battery of

20 \mathrm{~V}

is connected to the combination)

A

200 \,\mu \mathrm{C}

B 200 C

C

10 \,\mu \mathrm{C}

D 10 C

Correct Answer

Option A

Solution

Equivalent

C = \sum {{C_i}}

= 10\,\mu F

$\Rightarrow$ Charge

Q = CV

= 200\,\mu C

Q42

Two capacitors, each having capacitance

40 \,\mu \mathrm{F}

are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant

\mathrm{K}

such that the equivalence capacitance of the system became

24 \,\mu \mathrm{F}

. The value of

\mathrm{K}

will be :

A 1.5

B 2.5

C 1.2

D 3

Correct Answer

Option A

Solution

{{40K \times 40} \over {40K + 40}} = 24

40K = 24(K + 1)

40K = 24K + 24

16K = 24

K = {{24} \over {16}} = {3 \over 2} = 1.5

Q43

Which one of the following is the correct dimensional formula for the capacitance in F ?

\mathrm{M}, \mathrm{L}, \mathrm{T}

and

C

stand for unit of mass, length, time and charge,

A

[\mathrm{F}]=\left[\mathrm{CM}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]

B

[\mathrm{F}]=\left[\mathrm{C}^2 \mathrm{M}^{-2} \mathrm{~L}^2 \mathrm{~T}^2\right]

C

[\mathrm{F}]=\left[\mathrm{C}^2 \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]

D

[\mathrm{F}]=\left[\mathrm{CM}^{-2} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]

Correct Answer

Option C

Solution

The capacitance (in farads) is defined as the ratio of charge to potential difference.

Let's go through the steps: Capacitance is given by:

C = \frac{Q}{V}

where: $Q$ is the charge with dimensional symbol $C$ . $V$ is the potential difference.

Voltage (potential difference) is defined as energy per unit charge:

V = \frac{W}{Q}

where: $W$ is energy with dimensions:

[W] = [M L^2 T^{-2}]

Therefore, the dimensions of voltage are:

[V] = \frac{[ML^2T^{-2}]}{[C]}

Now, substitute this back into the expression for capacitance:

[C] = \frac{[Q]}{[ML^2T^{-2}]/[C]} = \frac{[C]^2}{ML^2T^{-2}}

Simplifying gives:

[C] = [C^2 M^{-1}L^{-2}T^2]

Comparing with the given options, we see that Option C is:

[F] = [C^2 M^{-1}L^{-2}T^2]

Thus, the correct dimensional formula for the capacitance in farads is given by Option C.

Q44

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

A decreases

B remains unchanged

C becomes infinite

D increases

Correct Answer

Option B

Solution

The capacitancce of parallel plate capacitor in which a metal plate of thickness

t

is inserted is given by

C = {{{\varepsilon _0}A} \over {d - t}}.\,\,\,\,\,

Here

t \to 0\,\,\,\,\,\,

$\therefore$

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

Q45

The work done in placing a charge of

8 \times {10^{ - 18}}

coulomb on a condenser of capacity

100

micro-farad is

A

16 \times {10^{ - 32}}\,\,joule

B

3.1 \times {10^{ - 26}}\,\,joule

C

4 \times {10^{ - 10}}\,\,joule

D

32 \times {10^{ - 32}}\,\,joule

Correct Answer

Option D

Solution

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by

U = {1 \over 2}{{{Q^2}} \over C}

= {1 \over 2} \times {{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}} \over {100 \times {{10}^{ - 6}}}}

= 32 \times {10^{ - 32}}J

Q46

Match with .

List - I		List - II
(a)	Capacitance, C	(i)	${M^1}{L^1}{T^{ - 3}}{A^{ - 1}}$
(b)	Permittivity of free space, ${\varepsilon _0}$	(ii)	${M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$
(c)	Permeability of free space, ${\mu _0}$	(iii)	${M^{ - 1}}{L^{ - 2}}{T^4}{A^2}$
(d)	Electric field, E	(iv)	${M^1}{L^1}{T^{ - 2}}{A^{ - 2}}$

A (a)

\to

(iii), (b)

\to

(ii), (c)

\to

(iv), (d)

\to

(i)

B (a)

\to

(iii), (b)

\to

(iv), (c)

\to

(ii), (d)

\to

(i)

C (a)

\to

(iv), (b)

\to

(ii), (c)

\to

(iii), (d)

\to

(i)

D (a)

\to

(iv), (b)

\to

(iii), (c)

\to

(ii), (d)

\to

(i)

Correct Answer

Option A

Solution

q = CV

[C] = \left[ {{q \over V}} \right] = {{(A \times T)} \over {M{L^2}{T^{ - 2}}}}

= {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}

[E] = \left[ {{F \over q}} \right] = {{ML{T^{ - 2}}} \over {AT}}

= ML{T^{ - 3}}{A^{ - 1}}

F = {{{q_1}{q_2}} \over {4\pi { \in _0}{r^2}}}

[{ \in _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}

Speed of light

c = {1 \over {\sqrt {{\mu _0}{ \in _0}} }}

{\mu _0} = {1 \over {{ \in _0}{c^2}}}

[{\mu _0}] = {1 \over {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]{{[L{T^{ - 1}}]}^2}}}

= [{M^1}{L^1}{T^{ - 2}}{A^{ - 2}}]

Q47

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.

Q48

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

Q49

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

Q50

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