# Electrostatic potential energy

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# We have an isolated conducting spherical shell of radius 10 cm. Some positive charge is given to it so that the resulting electric field has a maximum intensity of $1.8×{10}^{6}\text{\hspace{0.17em}\hspace{0.17em}}N{C}^{-1}.$ The same amount of negative charge is given to another isolated conducting spherical shell of radius 20 cm. Now, the first shell is placed inside the second so that both are concentric as shown in figure.

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## $1.8×{10}^{6}\text{\hspace{0.17em}}=\frac{9×{10}^{9}×{q}_{1}}{{\left(0.1\right)}^{2}}\text{\hspace{0.17em}\hspace{0.17em}}or\text{\hspace{0.17em}\hspace{0.17em}}{q}_{1}=2×{10}^{-6}\text{\hspace{0.17em}\hspace{0.17em}}C$$V=9×{10}^{9}×2×{10}^{-6}\left[\frac{1}{0.1}-\frac{1}{0.2}\right]=9×{10}^{4}\text{\hspace{0.17em}\hspace{0.17em}}V$$E=\frac{k{q}_{1}}{{r}^{2}}=\frac{9×{10}^{9}×2×{10}^{-6}}{{\left(0.20\right)}^{2}}=4.5×{10}^{5}\text{\hspace{0.17em}\hspace{0.17em}}N{C}^{-1}$Interaction energy is$U=\frac{kq}{0.2}\left(-q\right)=\frac{9×{10}^{9}{\left(2×{10}^{-6}\right)}^{2}}{0.2}=-0.18\text{\hspace{0.17em}\hspace{0.17em}}J$Self-energies are${U}_{1}=\frac{{q}_{2}}{8\pi {\epsilon }_{0}{r}_{1}}=\frac{9×{10}^{9}{\left(2×{10}^{-6}\right)}^{2}}{2×0.1}=0.18\text{\hspace{0.17em}\hspace{0.17em}}J$${U}_{2}=\frac{{q}_{2}}{8\pi {\epsilon }_{0}{r}_{2}}=\frac{9×{10}^{9}{\left(2×{10}^{-6}\right)}^{2}}{2×0.2}=0.09\text{\hspace{0.17em}\hspace{0.17em}}J$Total energy is $U+{U}_{1}+{U}_{2}=0.09\text{\hspace{0.17em}\hspace{0.17em}}J$Both the charges will get neutralized and there will be no charge left on any sphere. So no energy will be left in the system. It means whole amount of energy will convert into heat.

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