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There exists a non-conducting disc of radius R and having a charge density of $σ_{0}$ . What will be the work done by the electrostatic repulsion when radius of the disc becomes 2R. Assume during expansion charge distribution remains uniform at all the time. If work done by electric force is of the form $\frac{A {σ_{0}}^{2} R^{3}}{B \in_{0}}$ in simplest ratio. Then the value of $(A + B)$ will be?

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answer is 4.

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Detailed Solution

Self-energy of the disc is given by $U_{s e l f} \Rightarrow \frac{2 Q^{3}}{3 π^{2} \in_{0} R}, Q = σ_{0} . π R^{2}$

$W_{e} = U_{i} - U_{f} = \frac{2 Q^{2}}{3 π^{2} \in_{0}} [\frac{1}{R} - \frac{1}{2 R}] \Rightarrow \frac{Q^{2}}{3 π^{2} \in_{0} R} \Rightarrow \frac{σ^{2} π^{2} R^{4}}{3 π^{2} \in_{0} R} \Rightarrow \frac{σ^{2} R^{3}}{3 \in_{0}}$

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