Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is [figure]
- 3qQ4 0 a
- 3qQ8 0 a
- qQ4 0 a
- zero
Answer
(4) zero
Solution
Since CA = CB = 2a, the triangle is symmetric about the bisector of angle C, and D and E are corresponding points: CD = CE = a, BD = AE = a and AD = BE.
VD = q4 0 ( 1BD + 1CD + 1AD ) = q4 0 ( 2a + 1AD )
VE = q4 0 ( 1AE + 1CE + 1BE ) = q4 0 ( 2a + 1BE )
AD = BE VD = VE
W = Q(VE - VD) = 0
Answer: (4)