ElectrostaticsHard
Question
A system consisting of two charges 7μC and -2μC placed at (-9 cm, 0, 0) and (9 cm, 0, 0) respectively. This system of charges is placed in an external electric field E = 9r-2 kVm-1. The electrostatic energy of the configuration will be :-
Options
A.- 0.7 J
B.- 0.2 J
C.0.5 J
D.49.3 J
Solution
Point (-9cm, 0, 0) and (9cm, 0, 0) are situated at equipotential surface then
V(-9cm, 0, 0) = V(9cm, 0, 0) = -
E dr
= - 9 × 103
r-2dr = 105 vdt
The mutual interaction potential energy of two charges
=
= - 0.7 J
The potential energy of interaction of the two charges with the external electric field
= q1V(-9cm, 0, 0) + q2V(9cm, 0, 0)
= 7 × 10-6 × 105 + (-2 × 10-6) × 105 = 0.5J
Total electrostatic potential energy
= - 0.7 J + 0.5 J = - 0.2 J
V(-9cm, 0, 0) = V(9cm, 0, 0) = -
= - 9 × 103
The mutual interaction potential energy of two charges
=
The potential energy of interaction of the two charges with the external electric field
= q1V(-9cm, 0, 0) + q2V(9cm, 0, 0)
= 7 × 10-6 × 105 + (-2 × 10-6) × 105 = 0.5J
Total electrostatic potential energy
= - 0.7 J + 0.5 J = - 0.2 J
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