JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 5)
Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii $$\left( {{{{R_2}} \over {{R_1}}}} \right)$$ is :
$${n \over {n - 1}}$$
$${{2n} \over {2n + 1}}$$
$${{n + 1} \over n}$$
$${{2n + 1} \over n}$$
Explanation
Initially $$ = {C_0} = 4\pi {\varepsilon _0}{R_1}$$
Finally $${{4\pi {\varepsilon _0}{R_1}{R_2}} \over {{R_2} - {R_1}}} = n{C_0} = 4\pi {\varepsilon _0}n{R_1}$$
$$ \Rightarrow $$ $${{{R_2}} \over {{R_2} - {R_1}}} = n$$
$$ \Rightarrow $$$$1 - {{{R_1}} \over {{R_2}}} = {1 \over n}$$
$$ \Rightarrow $$ $${{{R_1}} \over {{R_2}}} = {{n - 1} \over n}$$
$$ \Rightarrow $$ $${{{R_2}} \over {{R_1}}} = {n \over {n - 1}}$$
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