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Two positive charges of magnitude q are placed at the ends of a side 1 of a square of side 2a . Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q movesfrom the middle of side 1 to the centre of square, its kinetic energy at the centre of square is(a) k2qQ(1-1/5^")/a (b) zero (c) k2qQ(1+1/5^)/a (d)k2qQ(1-2/5^)/a

Two positive charges of magnitude q are placed at the ends of a side 1 of a square of side 2a . Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q movesfrom the middle of side 1 to the centre of square, its kinetic energy at the centre of square is(a) k2qQ(1-1/5^")/a (b) zero (c) k2qQ(1+1/5^)/a (d)k2qQ(1-2/5^)/a

Grade:12th pass

1 Answers

Arun
25750 Points
6 years ago
Dear Nikita

Electric potential at P.
VP = (1/4πεo)q{1/a + 1/a - 1/(a√5) -1/(a√5)}
= (1/4πεoa)2q(1- 1/√5)
Electric potential at O
VO = (1/4πεo)q (1/√2 + 1/√2 - 1/√2 - 1/√2)
= 0
VP – VO =  (1/4πεoa)2q(1- 1/√5)
Total energy of the charge Q at point P = (1/4πεoa)2q(1- 1/√5)  here KE = 0
Total energy of the charge Q at point 0 = (1/4πεoa)2q(1- 1/√5) here PE = 0
Thus, KE = (1/4πεoa)2q(1- 1/√5)
Taking 1/4πεo = k
KE = k2q (1- 1/√5)/a
 = k2q (1- (1/5)1/2 )/a
Regards
Arun (askIITians forum expert)

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