2. Electric Potential and Capacitance
hard

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:

A

$\frac{ n }{ n -1}$

B

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

C

$\frac{ n +1}{ n }$

D

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

(JEE MAIN-2022)

Solution

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

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

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

Standard 12
Physics

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