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NEET 2011 (AIPMT, Mains) · Electrostatic Potential Energy

Electrostatic Potential Energy question from NEET 2011 (AIPMT, Mains)

Q56NEET 2011 · AIPMT · MAINSMedium

Electrostatic Potential Energy

Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2 $a$ . D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is

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Answer: (d)

$W = Q(V_E - V_D)$ .
With $BC = AC$ , the triangle is symmetric about the bisector of angle C, which swaps A with B and so D (mid BC) with E (mid CA).
Equal charges at the corners give $V_D = V_E$ , so $W = 0$ .
So the answer is (d) zero.

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