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# 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.mple 2.5 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.Suppose that the same system of charges is now placed in an external electric field $E = A (1/r^2); A = 9 \times 10^5 C m^{–2}$. What would the electrostatic energy of the configuration be?

$\begin{array}{1 1} 4.93\;J \\ 49.3\;J \\ 45.7\;J \\ 4.89\;J \end{array}$

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Solution :
$U= \large\frac{1}{4 \pi \in_0} \frac{q_1q_2}{r}$
$\quad= 9 \times 10^9 \times \large\frac{7 \times (-2) \times 10^{-12}}{0.18}$
$\qquad= -0.7\;J$
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_1V(r_1)+q_2V(r_2)=A$$\large\frac{\mu C}{0.09 \;m} +A \large\frac{-2 \mu C}{-0.09\;m}$
and the net electrostatic energy is
$\qquad= 70-20-0.7=49.3\;J$