Question

The drawing shows a square, each side of which has a length of L = 0.250 m. Two different positive charges q1 and q2 are fixed at the corners of the square. Find the electric potential energy of a third charge q3 = -4.00 x 10-9 C placed at corner A and then at corner B.

Answer 1

The electric potential at a point A due to charge q_1

V_1 = k*q_1 / r

here r = diagonal length of the square = sqrt(2) * a = sqrt(2) *(0.25 m) = 0.3535 m

vV_1 = (9*10^9 ) *(1.5*10^-9 C) / 0.3535 m = 38.19 V

The electric potential at a point A due to charge q_2

V_2 = k*q _2 / a == (9*10^9 ) *(4*10^-9 C) /0.25 m = 144 V

Total potential at a point A = V = V_1 + V_2 = 144 V + 38.19 V = 182.19 V

Potential energy = Vq = ( 182.19 V)(-4*10^-9 C) = - 728.8*10^-9 Joules

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The electric potential at a point B due to charge q_1

V_1 = k*q_1 / a

V_1 = (9*10^9 ) *(1.5*10^-9 C) / 0.25 m = 54 V

The electric potential at a point B due to charge q_2

V_2 = k*q _2 / r

here r = diagonal length of the square = sqrt(2) * a = sqrt(2) *(0.25 m) = 0.3535 m

V_2 = (9*10^9 ) *(4*10^-9 C) /0.3535 m = 101.83 V

Total potential at a point B = V = V_1 + V_2 = 54 V + 101.83 V = 155.83 V

potential energy = Vq =( 155.83 V)(-4*10^-9 C) = - 623.32 *10^-9 j