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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2022
  • Hard
  • MCQ

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 \(\left(\frac{ R _{2}}{ R _{1}}\right)\) is:

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Answer & Solution

Correct Answer

(A) \(\frac{ n }{ n -1}\)

Step-by-step Solution

Detailed explanation

Capacitance of isolated Conducting sphere \(=4 \pi \varepsilon_{0} R_{1}\)

By enclosing inside another sphere of radius \(R_{2}\), new capacitance \(=\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\)

Given: \(\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}=n \times 4 \pi \varepsilon_{0} R_{1}\)

\(\frac{ R _{2}}{\left( R _{2}- R _{1}\right)}= n \Rightarrow \frac{\frac{ R _{2}}{ R _{1}}}{\left(\frac{ R _{2}}{ R _{1}}-1\right)}= n\)

\(\frac{ R _{2}}{ R _{1}}= n \frac{ R _{2}}{ R _{1}}- n \Rightarrow \frac{ R _{2}}{ R _{1}}=\frac{ n }{( n -1)}\)

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