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# The capacity of an isolated conducting sphere of radius R is proportional toA. ${R^2}$B. $\dfrac{1}{{{R^2}}}$C. $\dfrac{1}{R}$D. $R$

Last updated date: 19th Jun 2024
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Hint: In this question, we need to determine the capacity (or capacitance) of an isolated conducting sphere of radius R. For this, we will use the relation between the capacitance value of the conducting sphere and the radius of the sphere.

Complete step by step answer:The potential of a charged conducting sphere is given as $V = \dfrac{Q}{{4\pi {\varepsilon _0}R}}$ where, V is the potential, Q is the charge on the sphere, R is the radius of the sphere and ${\varepsilon _0}$ is the permittivity of the free space.
The ratio of the charge of the sphere to the change in the potential difference of the sphere results in the capacity (or the capacitance) of the sphere. Mathematically, $C = \dfrac{Q}{{\vartriangle V}}$.
Let ‘a’ and ‘b’ be the inner and the outer radius of the sphere carrying a charge of ‘Q’ on the surface. Then, the change in the potential is given as:
$\vartriangle V = \dfrac{Q}{{\left[ {\dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right)}}} \right]}}$.
Substituting the value of $\vartriangle V = \dfrac{Q}{{\left[ {\dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right)}}} \right]}}$ in the equation $C = \dfrac{Q}{{\vartriangle V}}$ to determine the capacitance of the conducting sphere.
$C = \dfrac{Q}{{\vartriangle V}} \\ = \dfrac{Q}{{\dfrac{Q}{{\left[ {\dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right)}}} \right]}}}} \\ = \dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right)}} - - - - (i) \\$

According to the question, the conducting sphere is isolated, which implies that the inner radius is ‘R’ whereas the outer radius is tending towards infinity. So, substituting these values in the equation (i), we get
$C = \dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right)}} \\ = \dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{R} - \dfrac{1}{\infty }} \right)}} \\ = \dfrac{{4\pi {\varepsilon _0}}}{{\left( {\dfrac{1}{R} - 0} \right)}} \\ = 4\pi {\varepsilon _0}R \\$
Therefore, we can write $C \propto R$
Hence, the capacity of the isolated conducting sphere is proportional to R.

Option D is correct.

Note:Students must be careful while substituting the values of the inner and the outer radius of the isolated sphere. Isolated sphere means that we have to take the outer radius to be the radius of the earth which is very large as compared to the inner radius of the sphere.