Capacitor Questions — NEET Physics

Q1

Five capacitors of capacitances

C_1=C_2=C_3=C_4=10 \mu \mathrm{~F}

and

C_5=2.5 \mu \mathrm{~F}

are connected as shown, along with a battery of 50 V . The equivalent capacitance and the charges on each capacitor respectively are:

A

5 \mu \mathrm{~F}, 125 \mu \mathrm{C}

on

C_1

to

C_4

and

25 \mu \mathrm{C}

on

C_5

B

5 \mu \mathrm{~F}, 125 \mu \mathrm{C}

on all capacitors

C

5 \mu \mathrm{~F}, 250 \mu \mathrm{C}

on all capacitors

D

4 \mu \mathrm{~F}, 250 \mu \mathrm{C}

on

C_1

to

C_4

and

125 \mu \mathrm{C}

on

C_5

Correct Answer

Option B

Solution

\begin{aligned} & C_{\mathrm{eq}}=2.5+2.5=5 \mu \mathrm{~F}^ˋ \\ & q_1=q_2=q_3=q_4=2.5 \times 50=125 \mu \mathrm{C} \\ & q_5=2.5 \times 50=125 \mu \mathrm{C} \end{aligned}

Q2

Consider two uncharged capacitors of equal capacitance 200 pF . One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

A

0.5 \times 10^{-6} \mathrm{~J}

B 1.0 J

C

1.0 \times 10^{-6} \mathrm{~J}

D 0.5 J

Correct Answer

Option A

Solution

Energy loss $=\dfrac{1}{2} \dfrac{C_1 C_2}{C_1+C_2} V^2$

\begin{aligned} & =\frac{1}{2}\left(\frac{200 \times 200}{400}\right) \times 10^{-12}(100)^2 \\ & =\frac{1}{2} \times 10^6 \times 10^{-12}=0.5 \times 10^{-6} \mathrm{~J} \end{aligned}

Q3

The plates of a parallel plate capacitor are separated by

d

. Two slabs of different dielectric constant

K_1

and

K_2

with thickness

\dfrac{3}{8} d

and

\dfrac{d}{2}

, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If

K_1=1.25 K_2

, the value of

K_1

is:

A 1.60

B 1.33

C 2.66

D 2.33

Correct Answer

Option C

Solution

Using $C_{e q}=\dfrac{\varepsilon_0 A}{\dfrac{t_1}{K_1}+\dfrac{t_2}{K_2}+\dfrac{t_3}{K_3}}$ here $C_0=\dfrac{\varepsilon_0 A}{d}, t_1=\dfrac{3 d}{8}, t_2=\dfrac{d}{2}, t_3=\dfrac{d}{8}$

K_1=K_1, \quad K_2=\frac{K_1}{1.25} \text{ and } K_3=1

Given $C_{\text{eq }}=2 C_0$

\begin{aligned} & \Rightarrow 2 C_0=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d \times 1.25}{2 k_1}+\frac{d}{8}} \\ & \Rightarrow \frac{2 \varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d}{2 k_1} \times \frac{5}{4}+\frac{d}{8}} \\ & \Rightarrow 2=\frac{1}{\frac{3}{8 k_1}+\frac{5}{8 k_1}+\frac{1}{8}} \Rightarrow k_1=\frac{8}{3}=2.66 \end{aligned}

Q4

A

12 \mathrm{~pF}

capacitor is connected to a

50 \mathrm{~V}

battery, the electrostatic energy stored in the capacitor in

\mathrm{nJ}

is

A 15

B 7.5

C 0.3

D 150

Correct Answer

Option A

Solution

The electrostatic energy stored in a capacitor can be calculated using the formula:

U = \frac{1}{2} C V^2

Where:

U

is the stored energy in joules (J)

C

is the capacitance in farads (F)

V

is the voltage in volts (V) Given:

C = 12 \mathrm{~pF}

(which is

12 \times 10^{-12}

F)

V = 50 \mathrm{~V}

Substitute the values into the formula:

U = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2

Calculate the values inside the parentheses first:

50^2 = 2500

Now substitute this back into the equation:

U = \frac{1}{2} \times 12 \times 10^{-12} \times 2500

Perform the multiplication:

U = \frac{1}{2} \times 30000 \times 10^{-12}

Which simplifies to:

U = 15000 \times 10^{-12}

Convert this to nanojoules (nJ) by recognizing that

1 \mathrm{~nJ} = 10^{-9} \mathrm{~J}

:

U = 15 \mathrm{~nJ}

Thus, the electrostatic energy stored in the capacitor is: Option A: 15 nJ

Q5

The capacitance of a capacitor with charge

q

and a potential difference

V

depends on

A both

q

and

V

B the geometry of the capacitor

C

q

only

D

V

only

Correct Answer

Option B

Solution

The correct answer is Option B: the geometry of the capacitor .

Here's why: Capacitance (denoted by *C*) is a fundamental property of a capacitor that describes its ability to store electrical energy.

It's defined by the ratio of the charge stored on the capacitor ( q ) to the potential difference across its plates ( V ):

C = \frac{q}{V}

While the formula seems to suggest that capacitance depends on both charge and voltage, the reality is that the geometry of the capacitor determines its capacitance.

Here's a breakdown: Charge ( q ) : The charge stored on a capacitor is directly proportional to the applied voltage.

If you increase the voltage, you increase the charge stored.

However, the capacitance itself remains constant for a given capacitor.

Voltage ( V ) : Similarly, the voltage across a capacitor is directly proportional to the charge stored.

Increasing the charge increases the voltage, but again, the capacitance remains unchanged.

Geometry : The geometry of a capacitor dictates how much electric field is created between its plates for a given charge.

This electric field determines the potential difference between the plates.

Here are some key factors: Area of plates (A) : Larger plates can hold more charge for a given voltage, resulting in higher capacitance.

Distance between plates (d) : Smaller distances between plates create a stronger electric field, which leads to higher capacitance.

Dielectric material : The material between the plates (dielectric) affects the strength of the electric field and therefore the capacitance.

A material with a higher dielectric constant increases the capacitance.

In summary, while charge and voltage are related to capacitance through the formula, they are not the determining factors.

It's the capacitor's physical characteristics – its geometry – that ultimately determine its capacitance.

Q6

The steady state current in the circuit shown below is :

A 0.67 A

B 1.5 A

C 2 A

D 1 A

Correct Answer

Option C

Solution

At steady state, capacitor will be completely charged and will not allow current to pass through it.

The simplified circuit will be :

i=\frac{V}{R} \Rightarrow i=\frac{10}{2+3}=2 \mathrm{~A}

Q7

In the following circuit, the equivalent capacitance between terminal A and terminal B is :

A 2

\mu

F

B 1

\mu

F

C 0.5

\mu

F

D 4

\mu

F

Correct Answer

Option A

Solution

Given circuit is balanced Wheatstone bridge

\begin{aligned} C_{A B} & =1+1 \\ & =2 \mu F \end{aligned}

Q8

The equivalent capacitance of the arrangement shown in figure is:

A 30

\mu

F

B 15

\mu

F

C 25

\mu

F

D 20

\mu

F

Correct Answer

Option D

Solution

C_{eq}=5+15=20\mu F

Q9

To produce an instantaneous displacement current of

2 \mathrm{~mA}

in the space between the parallel plates of a capacitor of capacitance

4 ~\mu \mathrm{F}

, the rate of change of applied variable potential difference

\left(\dfrac{\mathrm{dV}}{\mathrm{dt}}\right)

must be :-

A

800 \mathrm{~V} / \mathrm{s}

B

500 \mathrm{~V} / \mathrm{s}

C

200 \mathrm{~V} / \mathrm{s}

D

400 \mathrm{~V} / \mathrm{s}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{Q}=\mathrm{CV} \\\\ & \begin{aligned} \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} \cdot \frac{\mathrm{dV}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}} & =\frac{\mathrm{I}}{\mathrm{C}}=\frac{2 \times 10^{-3}}{4 \times 10^{-6}} \\ & =\frac{10^3}{2}=500 \frac{\mathrm{V}}{\mathrm{s}} \end{aligned} \end{aligned}

Q10

The equivalent capacitance of the system shown in the following circuit is:

A 3

\mu

F

B 6

\mu

F

C 9

\mu

F

D 2

\mu

F

Correct Answer

Option D

Solution

\mathrm{C_{AB}=\frac{3\times6}{3+6}=2\mu F}