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$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \;\mu C$ and $-2\; \mu C$ (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 $E=A\left(1 / r^{2}\right) ; A=9 \times 10^{5} \;C m ^{-2} .$ What would the electrostatic energy of the configuration be?
Solution
$(a)$ $U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{7 \times(-2) \times 10^{-12}}{0.18}$$=-0.7\, J$
$(b)$ $W=U_{2}-U_{1}=0-U=0-(-0.7)$$=0.7 \,J$
$(c)$ The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
$q_{1} V\left( r _{1}\right)+q_{2} V\left( r _{2}\right)$$=A \frac{7\, \mu C }{0.09\, m }+A \frac{-2\, \mu C }{0.09 \,m }$
and the net electrostatic energy is
$q_{1} V\left( r _{1}\right)+q_{2} V\left( r _{2}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}$$=A \frac{7\, \mu C }{0.09 \,m }+A \frac{-2 \,\mu C }{0.09 \,m }-0.7 \,J$
$=70-20-0.7=49.3 \,J$
