Capacitance of an isolated conducting sphere | vedclass

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(A) The capacitance of an isolated conducting sphere of radius $R_{1}$ is given by $C_{1} = 4 \pi \varepsilon_{0} R_{1}$.When this sphere is enclosed

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

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(A) The capacitance of an isolated conducting sphere of radius $R_{1}$ is given by $C_{1} = 4 \pi \varepsilon_{0} R_{1}$.When this sphere is enclosed by a concentric conducting sphere of radius $R_{2}$ which is connected to the earth,the system forms a spherical capacitor. The new capacitance is $C_{2} = \frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{R_{2} - R_{1}}$.According to the problem,the new capacitance is $n$ times the original capacitance: $C_{2} = n C_{1}$.Substituting the expressions: $\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{R_{2} - R_{1}} = n (4 \pi \varepsilon_{0} R_{1})$.Simplifying the equation: $\frac{R_{2}}{R_{2} - R_{1}} = n$.Dividing the numerator and denominator of the left side by $R_{1}$: $\frac{R_{2}/R_{1}}{(R_{2}/R_{1}) - 1} = n$.Let $x = \frac{R_{2}}{R_{1}}$. Then $\frac{x}{x - 1} = n$.$x = n(x - 1) \Rightarrow x = nx - n \Rightarrow n = nx - x \Rightarrow n = x(n - 1)$.Therefore,$x = \frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$.

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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