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JEE Main 202225 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2022 · 25 July · Evening Shift · Physics · Capacitance

Capacitance of an isolated conducting sphere of radius R _ 1...

Capacitance of an isolated conducting sphere of radius R1{R}_{1} becomes nn times when it is enclosed by a concentric conducting sphere of radius R2{R}_{2} connected to earth. The ratio of their radii (R2R1)\left(\dfrac{{R}_{2}}{{R}_{1}}\right) is:

Options

  1. Ann−1\dfrac{n}{n - 1}
  2. B2n2n+1\dfrac{2n}{2n + 1}
  3. Cn+1n\dfrac{n + 1}{n}
  4. D2n+1n\dfrac{2n + 1}{n}

Correct answer

A. nn−1\dfrac{n}{n - 1}

Step-by-step solution

Capacitance of isolated conducting sphere C=4πε0R1C = 4\pi{ε}_{0}{R}_{1}

By enclosing inside another sphere of radius R2{R}_{2}, new capacitance C′=4πε0R1R2(R2−R1){C}^{'} = \dfrac{4\pi{ε}_{0}{R}_{1}{R}_{2}}{\left({R}_{2} - {R}_{1}\right)}

Given: 4πε0R1R2(R2−R1)=n×4πε0R1\dfrac{4\pi{ε}_{0}{R}_{1}{R}_{2}}{\left({R}_{2} - {R}_{1}\right)} = n \times 4\pi{ε}_{0}{R}_{1}

⇒R2(R2−R1)=n⇒R2R1(R2R1−1)=n⇒ \dfrac{{R}_{2}}{\left({R}_{2} - {R}_{1}\right)} = n ⇒ \dfrac{\dfrac{{R}_{2}}{{R}_{1}}}{\left(\dfrac{{R}_{2}}{{R}_{1}} - 1\right)} = n

⇒R2R1=nR2R1−n⇒R2R1=n(n−1)⇒ \dfrac{{R}_{2}}{{R}_{1}} = n\dfrac{{R}_{2}}{{R}_{1}} - n ⇒ \dfrac{{R}_{2}}{{R}_{1}} = \dfrac{n}{\left(n - 1\right)}

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