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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?

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1 Answer

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.
answered Jun 11, 2014 by meena.p
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