If the inner sphere of a spherical capacitor is earthed instead of the outer, show that the total capacitance is \(4 \pi v_{0} b^{2} /(b-a)\) where \(a

The formula for capacitance is given by \(C = \frac{Q}{V}\), where \(Q\) is the charge given and \(V\) is the potential difference applied. The potential difference between the two spheres can be written as \(V = \frac{Q}{4\pi v_{0} (1/a - 1/b)}\) where \(v_{0}\) is the permittivity of free space.

To find the capacitance, we rewrite the capacitance formula as \(C = \frac{Q}{V} = \frac{Q}{\frac{Q}{4\pi v_{0} (1/a - 1/b)}}\) and solve for C. After simplifying, we get the capacitance \(C = \frac{4\pi v_{0} b^2}{(b - a)}\).

Since the inner sphere is earthed, the total charge on it will be zero. We know that the potential of the inner sphere\( V_{a}\) can be written as \(\frac{Q_{a}}{4\pi v_{0}a}\), where \(Q_{a}\) is the charge on the inner sphere. Since \(V_{a}\) is zero due to the earthed connection, \(Q_{a} = 0\). Now let's find the charge distribution between inner and outer surfaces of the outer sphere. Using the capacitance formula, we can deduce that \(\frac{Q_{a}}{4\pi v_{0}a} - \frac{Q_2}{4\pi v_{0}b^{-}} = V_{ab}\) This implies that \( -\frac{Q_{outer}}{4\pi v_{0}b^{-}} = V_{ab}\) Let \(Q_{outer}\) be the charge on the outer surface of the outer sphere and \(Q_{inner}\) be the charge on the inner surface of the outer sphere. Since the total charge given is \(Q\), we can write: \(Q_{outer} + Q_{inner} = Q\) Now, as \(Q_{a} = 0\), we know that the charge on the inner surface of the outer sphere is equal in magnitude but opposite in sign, i.e., \(Q_{inner} = - Q_{a} = 0\). Therefore, the proportion of the charges on the outer and inner surfaces of the outer sphere would be: Outer surface: \(Q_{outer} = Q\) Inner surface: \(Q_{inner} = 0\)