Solution
The electrostatic self-energy of a charge distribution can be calculated by assembling it element by element. Let’s build a circular layer of radius $R$ by adding thin concentric rings.
At an intermediate stage, let the radius of the layer be $x$.
The potential at the circumference of this layer (where we will add the next ring) is given by the problem statement:
$$V(x) = 4\sigma x$$
Now, we bring a charge $dq$ to add a ring of thickness $dx$ and radius $x$.
The charge $dq$ is:
$$dq = \sigma \cdot dA = \sigma (2\pi x dx)$$
The work done to add this ring is: $$dW = V(x) \cdot dq$$ $$dW = (4\sigma x) \cdot (2\pi \sigma x dx)$$ $$dW = 8\pi \sigma^2 x^2 dx$$
To find the total energy for radius $R$, we integrate from $0$ to $R$: $$U = \int_0^R 8\pi \sigma^2 x^2 dx$$ $$U = 8\pi \sigma^2 \left[ \frac{x^3}{3} \right]_0^R$$ $$U = \frac{8}{3} \pi \sigma^2 R^3$$
Answer: (d) $\frac{8}{3} \pi \sigma^2 R^3$