Browse Questions

Suppose that the same system of charges is now placed in an external electric field $E = A (1/r^2); A = 9 × 10^5 Cm^{–2}.$ What would the electrostatic energy of the configuration be?

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, $q_1 V(r_1)+q_e V(r_2) =A \large\frac{7 \mu C}{0.09 m} +A \large\frac{-2 \mu C}{0.09 m}$
and the net electrostatic energy is
$q_1V(r_1)+q_2V(r_2)+\large\frac{q_1q_2}{4 \pi \in_0 r_{12} } $$=A \large\frac{7 \mu C}{0.09 m} + A \large\frac{-2 \mu C}{0.09 \;m}$$-0.7 J$
$\qquad= 70-20-0.7 =49.3 J$
Hence C is the correct answer.