# a. Determine the electrostatic potential energy of a system consisting of two charges $7\mu C\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}-2\mu C$ (and with no external field) placed at (-9 cm, 0,0) and (9 cm, 0, 0) respectively. b. How much work is required to separate the two charges infinitely away from each other? c. Suppose that the same system of charges is now placed in an external electric field $E=A\left(\frac{1}{{r}^{2}}\right),A=9×{10}^{5}C{m}^{-2}$. What would the electrostatic energy of the configuration be?Solution in Tamil

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## a. $U=\frac{1}{4\pi {\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{r}=9×{10}^{9}×\frac{7×\left(-2\right)×{10}^{-12}}{0.18}=-0.7J$ b. $W={U}_{2}-{U}_{1}=0-U=0\left(-0.7\right)=0.7J$ b. $W={U}_{2}-{U}_{1}=0-U=0-\left(0.7\right)=0.7J$ c. Energy of interaction of the two charges with the external electric field. ${q}_{1}V\left({r}_{1}\right)+{q}_{2}V\left({r}_{2}\right)=A\frac{7}{0.09}+A\frac{-2}{0.09}$ Net electrostatic energy is : ${q}_{1}V\left({r}_{1}\right)+{q}_{2}V\left({r}_{2}\right)+\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon }_{0}{r}_{12}}=A\frac{7}{0.09}+A\frac{-2}{0.09}-0.7=70-20-0.7=49.3J$

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