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Capacitance

165684 A parallel plate capacitor is charged. If the plates are pulled apart

1 the capacitance increases

2 the potential difference increases

3 the total charge increases

4 the charge and potential difference remain the same

Explanation:

: $E_{net} = E_{1} + E_{2}$
$= \frac{Q}{2 A ε_{0}} + \frac{Q}{2 A ε_{0}} = \frac{Q}{A ε_{0}}$
$V = E \times d$
$V = \frac{Q}{A ε_{0}} d$
$\frac{Q}{V} = \frac{A ε_{0}}{d} = C$
$C = \frac{Q}{V}$
$C = \frac{A ε_{0}}{d}$
$d =$ Constant distance Increases Capacitance decreases hence voltage should increases.
Hence, we can say that potential difference will be Increases.

[DCE-2009]

Capacitance

165685 A parallel plate capacitor of capacitance $2 F$ is charged to a potential $V$ . 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 $E_{2}$ . The ratio $E_{2} / E_{1}$ is:

1

2 : 1

2

1 : 2

3

1 : 4

4

2 : 3

Explanation:

: Given that, capacitor of capacitance $= 2 F$ at Initially, potential $= V$
Charge of capacitor
$Q_{1} = CV = (2) V$
Energy stored in capacitor,
$E_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} (2) V^{2} = V^{2}$
at Finally,
Charge on each capacitor, $Q_{2} = \frac{Q_{1}}{2} = \frac{2 V}{2} = V$
Then, energy stored finally in two capacitors
$E_{2} = 2 (\frac{1}{2} \frac{Q_{2}^{2}}{C}) = \frac{V^{2}}{2}$
Equation (ii) divided by equation (i)
$\frac{E_{2}}{E_{1}} = \frac{1}{2}$

[JEE Main-11.04.2023

Capacitance

165686 A $40 μ F$ capacitor in a defibrillator is charged to $3000 V$ . The energy stored in the capacitor is set through the patient during a pulse of duration $2 ms$ . The power delivered to the patient is :

1

45 kW

2

90 kW

3

180 kW

4

360 kW

Explanation:

: Given,
$C = 40 μ F$
$V = 3000 V$
$t = 2 ms = 2 \times 10^{- 3} s$
The energy store in the capacitor,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 40 \times (3000)^{2} \times 10^{- 6}$
$= 180 J$
The power delivered in $2 ms$ ,
$P = \frac{Energy delivered}{Time} = \frac{180 J}{2 \times 10^{- 3} s}$
$= 90 \times 10^{3} W$
$= 90 kW$

Capacitance

165688 Two condensers, one of capacity $C$ and the other of capacity $\frac{C}{2}$ , are connected to a V-volt battery, as shown.

The work done in charging fully both the condensers is

1

{CV}^{2}

2

\frac{1}{4} {CV}^{2}

3

\frac{3}{4} {CV}^{2}

4

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

Explanation:

: Energy stored in a condenser,
$U = \frac{1}{2} {CV}^{2}$
For first condenser-
$U_{1} = \frac{1}{2} {CV}^{2} (capacity = C)$
For second condenser-
$U_{2} = \frac{1}{2} (\frac{C}{2}) \cdot V^{2}$
$U_{2} = \frac{1}{4} {CV}^{2}$
$(capactity = \frac{C}{2})$
So, Work done $=$ Energy stored in condensers.
$W = U_{1} + U_{2}$
$= \frac{1}{2} {CV}^{2} + \frac{1}{4} {CV}^{2}$
$= \frac{3}{4} {CV}^{2}$

[JIPMER-2017

Capacitance

165689 Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $(\frac{R_{2}}{R_{1}})$ is :

1

\frac{n}{n - 1}

2

\frac{2 n}{2 n + 1}

3

\frac{n + 1}{n}

4

\frac{2 n + 1}{n}

Explanation:

: Given, $C_{2} = {nC}_{1}$

For sphere (i),
Capacitance $(C_{1}) = 4 π ε_{0} R_{1}$
For sphere (ii),
$Capacitance (C_{2}) = 4 π ε_{0} \cdot (\frac{R_{1} R_{2}}{R_{2} - R_{1}})$
According to question,
$C_{2} = {nC}_{1}$
$4 π ε_{0} (\frac{R_{1} R_{2}}{R_{2} - R_{1}}) = n 4 π ε_{0} R_{1}$
$\frac{R_{1} R_{2}}{R_{2} - R_{1}} = {nR}_{1}$
$\frac{\frac{R_{2}}{R_{1}}}{\frac{R_{2}}{R_{1}} - 1} = n$
$\frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

[JEE Main-25.07.2022

Capacitance

165684 A parallel plate capacitor is charged. If the plates are pulled apart

1 the capacitance increases

2 the potential difference increases

3 the total charge increases

4 the charge and potential difference remain the same

Explanation:

: $E_{net} = E_{1} + E_{2}$
$= \frac{Q}{2 A ε_{0}} + \frac{Q}{2 A ε_{0}} = \frac{Q}{A ε_{0}}$
$V = E \times d$
$V = \frac{Q}{A ε_{0}} d$
$\frac{Q}{V} = \frac{A ε_{0}}{d} = C$
$C = \frac{Q}{V}$
$C = \frac{A ε_{0}}{d}$
$d =$ Constant distance Increases Capacitance decreases hence voltage should increases.
Hence, we can say that potential difference will be Increases.

[DCE-2009]

Capacitance

165685 A parallel plate capacitor of capacitance $2 F$ is charged to a potential $V$ . 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 $E_{2}$ . The ratio $E_{2} / E_{1}$ is:

1

2 : 1

2

1 : 2

3

1 : 4

4

2 : 3

Explanation:

: Given that, capacitor of capacitance $= 2 F$ at Initially, potential $= V$
Charge of capacitor
$Q_{1} = CV = (2) V$
Energy stored in capacitor,
$E_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} (2) V^{2} = V^{2}$
at Finally,
Charge on each capacitor, $Q_{2} = \frac{Q_{1}}{2} = \frac{2 V}{2} = V$
Then, energy stored finally in two capacitors
$E_{2} = 2 (\frac{1}{2} \frac{Q_{2}^{2}}{C}) = \frac{V^{2}}{2}$
Equation (ii) divided by equation (i)
$\frac{E_{2}}{E_{1}} = \frac{1}{2}$

[JEE Main-11.04.2023

Capacitance

165686 A $40 μ F$ capacitor in a defibrillator is charged to $3000 V$ . The energy stored in the capacitor is set through the patient during a pulse of duration $2 ms$ . The power delivered to the patient is :

1

45 kW

2

90 kW

3

180 kW

4

360 kW

Explanation:

: Given,
$C = 40 μ F$
$V = 3000 V$
$t = 2 ms = 2 \times 10^{- 3} s$
The energy store in the capacitor,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 40 \times (3000)^{2} \times 10^{- 6}$
$= 180 J$
The power delivered in $2 ms$ ,
$P = \frac{Energy delivered}{Time} = \frac{180 J}{2 \times 10^{- 3} s}$
$= 90 \times 10^{3} W$
$= 90 kW$

Capacitance

165688 Two condensers, one of capacity $C$ and the other of capacity $\frac{C}{2}$ , are connected to a V-volt battery, as shown.

The work done in charging fully both the condensers is

1

{CV}^{2}

2

\frac{1}{4} {CV}^{2}

3

\frac{3}{4} {CV}^{2}

4

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

Explanation:

: Energy stored in a condenser,
$U = \frac{1}{2} {CV}^{2}$
For first condenser-
$U_{1} = \frac{1}{2} {CV}^{2} (capacity = C)$
For second condenser-
$U_{2} = \frac{1}{2} (\frac{C}{2}) \cdot V^{2}$
$U_{2} = \frac{1}{4} {CV}^{2}$
$(capactity = \frac{C}{2})$
So, Work done $=$ Energy stored in condensers.
$W = U_{1} + U_{2}$
$= \frac{1}{2} {CV}^{2} + \frac{1}{4} {CV}^{2}$
$= \frac{3}{4} {CV}^{2}$

[JIPMER-2017

Capacitance

165689 Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $(\frac{R_{2}}{R_{1}})$ is :

1

\frac{n}{n - 1}

2

\frac{2 n}{2 n + 1}

3

\frac{n + 1}{n}

4

\frac{2 n + 1}{n}

Explanation:

: Given, $C_{2} = {nC}_{1}$

For sphere (i),
Capacitance $(C_{1}) = 4 π ε_{0} R_{1}$
For sphere (ii),
$Capacitance (C_{2}) = 4 π ε_{0} \cdot (\frac{R_{1} R_{2}}{R_{2} - R_{1}})$
According to question,
$C_{2} = {nC}_{1}$
$4 π ε_{0} (\frac{R_{1} R_{2}}{R_{2} - R_{1}}) = n 4 π ε_{0} R_{1}$
$\frac{R_{1} R_{2}}{R_{2} - R_{1}} = {nR}_{1}$
$\frac{\frac{R_{2}}{R_{1}}}{\frac{R_{2}}{R_{1}} - 1} = n$
$\frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

[JEE Main-25.07.2022

Capacitance

165684 A parallel plate capacitor is charged. If the plates are pulled apart

1 the capacitance increases

2 the potential difference increases

3 the total charge increases

4 the charge and potential difference remain the same

Explanation:

: $E_{net} = E_{1} + E_{2}$
$= \frac{Q}{2 A ε_{0}} + \frac{Q}{2 A ε_{0}} = \frac{Q}{A ε_{0}}$
$V = E \times d$
$V = \frac{Q}{A ε_{0}} d$
$\frac{Q}{V} = \frac{A ε_{0}}{d} = C$
$C = \frac{Q}{V}$
$C = \frac{A ε_{0}}{d}$
$d =$ Constant distance Increases Capacitance decreases hence voltage should increases.
Hence, we can say that potential difference will be Increases.

[DCE-2009]

Capacitance

165685 A parallel plate capacitor of capacitance $2 F$ is charged to a potential $V$ . 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 $E_{2}$ . The ratio $E_{2} / E_{1}$ is:

1

2 : 1

2

1 : 2

3

1 : 4

4

2 : 3

Explanation:

: Given that, capacitor of capacitance $= 2 F$ at Initially, potential $= V$
Charge of capacitor
$Q_{1} = CV = (2) V$
Energy stored in capacitor,
$E_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} (2) V^{2} = V^{2}$
at Finally,
Charge on each capacitor, $Q_{2} = \frac{Q_{1}}{2} = \frac{2 V}{2} = V$
Then, energy stored finally in two capacitors
$E_{2} = 2 (\frac{1}{2} \frac{Q_{2}^{2}}{C}) = \frac{V^{2}}{2}$
Equation (ii) divided by equation (i)
$\frac{E_{2}}{E_{1}} = \frac{1}{2}$

[JEE Main-11.04.2023

Capacitance

165686 A $40 μ F$ capacitor in a defibrillator is charged to $3000 V$ . The energy stored in the capacitor is set through the patient during a pulse of duration $2 ms$ . The power delivered to the patient is :

1

45 kW

2

90 kW

3

180 kW

4

360 kW

Explanation:

: Given,
$C = 40 μ F$
$V = 3000 V$
$t = 2 ms = 2 \times 10^{- 3} s$
The energy store in the capacitor,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 40 \times (3000)^{2} \times 10^{- 6}$
$= 180 J$
The power delivered in $2 ms$ ,
$P = \frac{Energy delivered}{Time} = \frac{180 J}{2 \times 10^{- 3} s}$
$= 90 \times 10^{3} W$
$= 90 kW$

Capacitance

165688 Two condensers, one of capacity $C$ and the other of capacity $\frac{C}{2}$ , are connected to a V-volt battery, as shown.

The work done in charging fully both the condensers is

1

{CV}^{2}

2

\frac{1}{4} {CV}^{2}

3

\frac{3}{4} {CV}^{2}

4

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

Explanation:

: Energy stored in a condenser,
$U = \frac{1}{2} {CV}^{2}$
For first condenser-
$U_{1} = \frac{1}{2} {CV}^{2} (capacity = C)$
For second condenser-
$U_{2} = \frac{1}{2} (\frac{C}{2}) \cdot V^{2}$
$U_{2} = \frac{1}{4} {CV}^{2}$
$(capactity = \frac{C}{2})$
So, Work done $=$ Energy stored in condensers.
$W = U_{1} + U_{2}$
$= \frac{1}{2} {CV}^{2} + \frac{1}{4} {CV}^{2}$
$= \frac{3}{4} {CV}^{2}$

[JIPMER-2017

Capacitance

165689 Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $(\frac{R_{2}}{R_{1}})$ is :

1

\frac{n}{n - 1}

2

\frac{2 n}{2 n + 1}

3

\frac{n + 1}{n}

4

\frac{2 n + 1}{n}

Explanation:

: Given, $C_{2} = {nC}_{1}$

For sphere (i),
Capacitance $(C_{1}) = 4 π ε_{0} R_{1}$
For sphere (ii),
$Capacitance (C_{2}) = 4 π ε_{0} \cdot (\frac{R_{1} R_{2}}{R_{2} - R_{1}})$
According to question,
$C_{2} = {nC}_{1}$
$4 π ε_{0} (\frac{R_{1} R_{2}}{R_{2} - R_{1}}) = n 4 π ε_{0} R_{1}$
$\frac{R_{1} R_{2}}{R_{2} - R_{1}} = {nR}_{1}$
$\frac{\frac{R_{2}}{R_{1}}}{\frac{R_{2}}{R_{1}} - 1} = n$
$\frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

[JEE Main-25.07.2022

Capacitance

165684 A parallel plate capacitor is charged. If the plates are pulled apart

1 the capacitance increases

2 the potential difference increases

3 the total charge increases

4 the charge and potential difference remain the same

Explanation:

: $E_{net} = E_{1} + E_{2}$
$= \frac{Q}{2 A ε_{0}} + \frac{Q}{2 A ε_{0}} = \frac{Q}{A ε_{0}}$
$V = E \times d$
$V = \frac{Q}{A ε_{0}} d$
$\frac{Q}{V} = \frac{A ε_{0}}{d} = C$
$C = \frac{Q}{V}$
$C = \frac{A ε_{0}}{d}$
$d =$ Constant distance Increases Capacitance decreases hence voltage should increases.
Hence, we can say that potential difference will be Increases.

[DCE-2009]

Capacitance

165685 A parallel plate capacitor of capacitance $2 F$ is charged to a potential $V$ . 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 $E_{2}$ . The ratio $E_{2} / E_{1}$ is:

1

2 : 1

2

1 : 2

3

1 : 4

4

2 : 3

Explanation:

: Given that, capacitor of capacitance $= 2 F$ at Initially, potential $= V$
Charge of capacitor
$Q_{1} = CV = (2) V$
Energy stored in capacitor,
$E_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} (2) V^{2} = V^{2}$
at Finally,
Charge on each capacitor, $Q_{2} = \frac{Q_{1}}{2} = \frac{2 V}{2} = V$
Then, energy stored finally in two capacitors
$E_{2} = 2 (\frac{1}{2} \frac{Q_{2}^{2}}{C}) = \frac{V^{2}}{2}$
Equation (ii) divided by equation (i)
$\frac{E_{2}}{E_{1}} = \frac{1}{2}$

[JEE Main-11.04.2023

Capacitance

165686 A $40 μ F$ capacitor in a defibrillator is charged to $3000 V$ . The energy stored in the capacitor is set through the patient during a pulse of duration $2 ms$ . The power delivered to the patient is :

1

45 kW

2

90 kW

3

180 kW

4

360 kW

Explanation:

: Given,
$C = 40 μ F$
$V = 3000 V$
$t = 2 ms = 2 \times 10^{- 3} s$
The energy store in the capacitor,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 40 \times (3000)^{2} \times 10^{- 6}$
$= 180 J$
The power delivered in $2 ms$ ,
$P = \frac{Energy delivered}{Time} = \frac{180 J}{2 \times 10^{- 3} s}$
$= 90 \times 10^{3} W$
$= 90 kW$

Capacitance

165688 Two condensers, one of capacity $C$ and the other of capacity $\frac{C}{2}$ , are connected to a V-volt battery, as shown.

The work done in charging fully both the condensers is

1

{CV}^{2}

2

\frac{1}{4} {CV}^{2}

3

\frac{3}{4} {CV}^{2}

4

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

Explanation:

: Energy stored in a condenser,
$U = \frac{1}{2} {CV}^{2}$
For first condenser-
$U_{1} = \frac{1}{2} {CV}^{2} (capacity = C)$
For second condenser-
$U_{2} = \frac{1}{2} (\frac{C}{2}) \cdot V^{2}$
$U_{2} = \frac{1}{4} {CV}^{2}$
$(capactity = \frac{C}{2})$
So, Work done $=$ Energy stored in condensers.
$W = U_{1} + U_{2}$
$= \frac{1}{2} {CV}^{2} + \frac{1}{4} {CV}^{2}$
$= \frac{3}{4} {CV}^{2}$

[JIPMER-2017

Capacitance

165689 Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $(\frac{R_{2}}{R_{1}})$ is :

1

\frac{n}{n - 1}

2

\frac{2 n}{2 n + 1}

3

\frac{n + 1}{n}

4

\frac{2 n + 1}{n}

Explanation:

: Given, $C_{2} = {nC}_{1}$

For sphere (i),
Capacitance $(C_{1}) = 4 π ε_{0} R_{1}$
For sphere (ii),
$Capacitance (C_{2}) = 4 π ε_{0} \cdot (\frac{R_{1} R_{2}}{R_{2} - R_{1}})$
According to question,
$C_{2} = {nC}_{1}$
$4 π ε_{0} (\frac{R_{1} R_{2}}{R_{2} - R_{1}}) = n 4 π ε_{0} R_{1}$
$\frac{R_{1} R_{2}}{R_{2} - R_{1}} = {nR}_{1}$
$\frac{\frac{R_{2}}{R_{1}}}{\frac{R_{2}}{R_{1}} - 1} = n$
$\frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

[JEE Main-25.07.2022

Capacitance

165684 A parallel plate capacitor is charged. If the plates are pulled apart

1 the capacitance increases

2 the potential difference increases

3 the total charge increases

4 the charge and potential difference remain the same

Explanation:

: $E_{net} = E_{1} + E_{2}$
$= \frac{Q}{2 A ε_{0}} + \frac{Q}{2 A ε_{0}} = \frac{Q}{A ε_{0}}$
$V = E \times d$
$V = \frac{Q}{A ε_{0}} d$
$\frac{Q}{V} = \frac{A ε_{0}}{d} = C$
$C = \frac{Q}{V}$
$C = \frac{A ε_{0}}{d}$
$d =$ Constant distance Increases Capacitance decreases hence voltage should increases.
Hence, we can say that potential difference will be Increases.

[DCE-2009]

Capacitance

165685 A parallel plate capacitor of capacitance $2 F$ is charged to a potential $V$ . 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 $E_{2}$ . The ratio $E_{2} / E_{1}$ is:

1

2 : 1

2

1 : 2

3

1 : 4

4

2 : 3

Explanation:

: Given that, capacitor of capacitance $= 2 F$ at Initially, potential $= V$
Charge of capacitor
$Q_{1} = CV = (2) V$
Energy stored in capacitor,
$E_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} (2) V^{2} = V^{2}$
at Finally,
Charge on each capacitor, $Q_{2} = \frac{Q_{1}}{2} = \frac{2 V}{2} = V$
Then, energy stored finally in two capacitors
$E_{2} = 2 (\frac{1}{2} \frac{Q_{2}^{2}}{C}) = \frac{V^{2}}{2}$
Equation (ii) divided by equation (i)
$\frac{E_{2}}{E_{1}} = \frac{1}{2}$

[JEE Main-11.04.2023

Capacitance

165686 A $40 μ F$ capacitor in a defibrillator is charged to $3000 V$ . The energy stored in the capacitor is set through the patient during a pulse of duration $2 ms$ . The power delivered to the patient is :

1

45 kW

2

90 kW

3

180 kW

4

360 kW

Explanation:

: Given,
$C = 40 μ F$
$V = 3000 V$
$t = 2 ms = 2 \times 10^{- 3} s$
The energy store in the capacitor,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 40 \times (3000)^{2} \times 10^{- 6}$
$= 180 J$
The power delivered in $2 ms$ ,
$P = \frac{Energy delivered}{Time} = \frac{180 J}{2 \times 10^{- 3} s}$
$= 90 \times 10^{3} W$
$= 90 kW$

Capacitance

165688 Two condensers, one of capacity $C$ and the other of capacity $\frac{C}{2}$ , are connected to a V-volt battery, as shown.

The work done in charging fully both the condensers is

1

{CV}^{2}

2

\frac{1}{4} {CV}^{2}

3

\frac{3}{4} {CV}^{2}

4

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

Explanation:

: Energy stored in a condenser,
$U = \frac{1}{2} {CV}^{2}$
For first condenser-
$U_{1} = \frac{1}{2} {CV}^{2} (capacity = C)$
For second condenser-
$U_{2} = \frac{1}{2} (\frac{C}{2}) \cdot V^{2}$
$U_{2} = \frac{1}{4} {CV}^{2}$
$(capactity = \frac{C}{2})$
So, Work done $=$ Energy stored in condensers.
$W = U_{1} + U_{2}$
$= \frac{1}{2} {CV}^{2} + \frac{1}{4} {CV}^{2}$
$= \frac{3}{4} {CV}^{2}$

[JIPMER-2017

Capacitance

165689 Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $(\frac{R_{2}}{R_{1}})$ is :

1

\frac{n}{n - 1}

2

\frac{2 n}{2 n + 1}

3

\frac{n + 1}{n}

4

\frac{2 n + 1}{n}

Explanation:

: Given, $C_{2} = {nC}_{1}$

For sphere (i),
Capacitance $(C_{1}) = 4 π ε_{0} R_{1}$
For sphere (ii),
$Capacitance (C_{2}) = 4 π ε_{0} \cdot (\frac{R_{1} R_{2}}{R_{2} - R_{1}})$
According to question,
$C_{2} = {nC}_{1}$
$4 π ε_{0} (\frac{R_{1} R_{2}}{R_{2} - R_{1}}) = n 4 π ε_{0} R_{1}$
$\frac{R_{1} R_{2}}{R_{2} - R_{1}} = {nR}_{1}$
$\frac{\frac{R_{2}}{R_{1}}}{\frac{R_{2}}{R_{1}} - 1} = n$
$\frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

[JEE Main-25.07.2022

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