Question

Two identical point charges (q = +2.90 x 10-6 C) are fixed at opposite corners of a square whose sides have a length of 0.540 m. A test charge (q0 = -1.80 x 10-8 C), with a mass of 9.30 x 10-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square.

Answer 1

potential at corner point

V1 = k q1 / r + k q2 / r

V1 = 9*10^9* 10^-6 / ( 0.54) * ( 2.9 + 2.9)

V1 = 9.67*10^4 V

potential at mid point

V2 = k (q1 + q2) / ( a/ sqrt(2) )

V2 = 9*10^9* 10^-6*2*2.9 / (.0.54/ sqrt (2) )

V2 = 13.671 *10^4 V

using energy conservation

0.5* m v^2 = q ( V2 - V1)

0.5* 9.3*10^-8* v^2 = 1.8*10^-8 * ( 13.671 - 9.67)*10^4

v = 124.45 m/s

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