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:
Pick an option to check your answer.
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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