(a) Assumption of Uniform Charge Distribution

No, strictly speaking, we cannot assume that the charge on the bulge is distributed uniformly.

In electrostatics, for a charged conductor, the surface charge density $\sigma$ is inversely related to the local radius of curvature. Since the radius of the bulge $r$ is much smaller than the radius of the sphere $R$ ($r \ll R$), the charge density on the curved surface of the bulge will be significantly higher than on the flatter parts of the large sphere.

However, for the purpose of the estimation requested in part (b), we will proceed by assuming a uniform charge density equal to that of the main sphere.

(b) Estimation of Charge $q$ on the Bulge

Let us assume the surface charge density $\sigma$ is uniform over the entire conductor. The total charge $Q$ is distributed over the sphere of radius $R$.

$$ \sigma = \frac{\text{Total Charge}}{\text{Surface Area}} = \frac{Q}{4\pi R^2} $$

We treat the bulge as a hemisphere of radius $r$. The surface area of this hemispherical bulge is: $$ A_{bulge} = 2\pi r^2 $$

The estimated charge $q$ on this area is: $$ q = \sigma \times A_{bulge} $$

Substituting the value of $\sigma$: $$ q = \left( \frac{Q}{4\pi R^2} \right) \times (2\pi r^2) $$

(c) Change in Capacitance $\Delta C$

The capacitance of the original sphere is given by: $$ C_{original} = 4\pi \epsilon_0 R $$

To find the new capacitance, we consider the effective total charge that the system can hold at the same potential $V$. The potential of the sphere is dominated by its main radius $R$: $$ V = \frac{Q}{4\pi \epsilon_0 R} $$

Due to the bulge, we have an additional charge $q$ (as calculated above) contributing to the capacity of the system. We can model the new capacitance $C_{new}$ as: $$ C_{new} = \frac{Q + q}{V} $$

Substituting the expressions for $q$ and $V$: $$ C_{new} = \frac{Q + \frac{Q r^2}{2 R^2}}{\left( \frac{Q}{4\pi \epsilon_0 R} \right)} $$

Simplifying the expression: $$ C_{new} = \frac{Q \left( 1 + \frac{r^2}{2 R^2} \right)}{\frac{Q}{4\pi \epsilon_0 R}} $$ $$ C_{new} = 4\pi \epsilon_0 R \left( 1 + \frac{r^2}{2 R^2} \right) $$ $$ C_{new} = 4\pi \epsilon_0 R + 4\pi \epsilon_0 R \left( \frac{r^2}{2 R^2} \right) $$ $$ C_{new} = 4\pi \epsilon_0 R + \frac{2\pi \epsilon_0 r^2}{R} $$

The change in capacitance $\Delta C$ is: $$ \Delta C = C_{new} – C_{original} $$ $$ \Delta C = \left( 4\pi \epsilon_0 R + \frac{2\pi \epsilon_0 r^2}{R} \right) – 4\pi \epsilon_0 R $$