Question

# A spherical shell of radius $$R_{1}$$ with a uniform charge $$q$$ has a point charge $$q_{0}$$ at its centre. Find the work performed by the electric forces during the shell expansion from radius $$R_{1}$$ to radius $$R_{2}$$.

Solution

## Initially, energy of the system, $$U_{i} = W_{1} + W_{12}$$ where, $$W_{1}$$ is the self energy and $$W_{12}$$ is the mutual energy.So, $$U_{i} = \dfrac {1}{2} \dfrac {q^{2}}{4\pi \epsilon_{0} R_{1}} + \dfrac {qq_{0}}{4\pi \epsilon_{0} R_{1}}$$and on expansion, energy of the system,$$U_{f} = W'_{1} + W'_{12}$$$$= \dfrac {1}{2} \dfrac {q^{2}}{4\pi \epsilon_{0}R_{2}} + \dfrac {qq_{0}}{4\pi \epsilon_{0} R_{2}}$$Now, work done by the field force, $$A$$ equals the decrement in the electrical energy,i.e. $$A = (U_{i} - U_{f}) = \dfrac {q(q_{0} + q/2)}{4\pi \epsilon_{0}} \left (\dfrac {1}{R_{1}} - \dfrac {1}{R_{2}}\right )$$Alternate : The work of electric forces is equal to the decrease in electric energy of the system,$$A = U_{i} - U_{f}$$In order to find the difference $$U_{i} - U_{f}$$, we note that upon expansion of the shell, the electric field and hence the energy localized in it, changed only in the hatched spherical layer consequently (Fig.).$$U_{i} - U_{f} = \int_{R_{1}}^{R_{2}} \dfrac {\epsilon_{0}}{2} (E_{1}^{2} - E_{2}^{2}) \cdot 4\pi r^{2} dr$$where $$E_{1}$$ and $$E_{2}$$ are the field intensities (in the hatched region at a distance $$r$$ from the centre of the system) before and after the expansion of the shell. By using Gauss' theorem, we find$$E_{1} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q + q_{0}}{r^{2}}$$ and $$E_{2} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{0}}{r^{2}}$$As a result of integration, we obtain$$A = \dfrac {q(q_{0} + q/2)}{4\pi \epsilon_{0}} \left (\dfrac {1}{R_{1}} - \dfrac {1}{R_{2}}\right )$$.Physics

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