Capacitors

JEE Main level, single correct, medium

14 of 180

Question 14 of 180

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:

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Answer: (A) nn−1\dfrac{n}{n-1}

Capacitance of isolated conducting sphere C=4πϵ0R1C=4\pi {\epsilon }_{0}{R}_{1}
By enclosing inside another sphere of radius R2{R}_{2}, new capacitance C′=4πϵ0R1R2(R2−R1){C}^{'}=\dfrac{4\pi {\epsilon }_{0}{R}_{1}{R}_{2}}{\left({R}_{2}-{R}_{1}\right)}
Given: 4πϵ0R1R2(R2−R1)=n×4πϵ0R1\dfrac{4\pi {\epsilon }_{0}{R}_{1}{R}_{2}}{\left({R}_{2}-{R}_{1}\right)}=n\times 4\pi {\epsilon }_{0}{R}_{1}
⇒R2(R2−R1)=n⇒R2R1(R2R1−1)=n\Rightarrow \dfrac{{R}_{2}}{\left({R}_{2}-{R}_{1}\right)}=n\Rightarrow \dfrac{\dfrac{{R}_{2}}{{R}_{1}}}{\left(\dfrac{{R}_{2}}{{R}_{1}}-1\right)}=n
⇒R2R1=nR2R1−n⇒R2R1=n(n−1)\Rightarrow \dfrac{{R}_{2}}{{R}_{1}}=n\dfrac{{R}_{2}}{{R}_{1}}-n \Rightarrow \dfrac{{R}_{2}}{{R}_{1}}=\dfrac{n}{\left(n-1\right)}

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