Question

The drawing shows a square, each side of which has a length of L = 0.25 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 = -7.00 10-9 C placed at corner A and then at corner B.

Answer 1

diagonal of the square = .25 * √(2)

√(2) is the square root of 2 (which equals 1.414)

corner A:
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k*q1/0.35 =(9*10^9)*(1.5*10^-9)/0.35 =38.6 V
k*q2/0.25 =(9*10^9)*(4*10^-9)/0.25 =144 V
then we add these to get the sum
38.6+144 = 182.6 V.

PE = q3 * (sum of V) = (-7.00*10^-9)* 182.6 = -1.2782*10^-6 J

Corner B:
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k*q1/0.25 =(9*10^9)*(1.5*10^-9)/0.25 =54 V
k*q2/0.35 =(9*10^9)*(4*10^-9)/0.35 =102.9 V
then we add these to get the sum
54+102.9 = 156.9 V.

PE = q3 * (sum of V) = (-7.00*10^-9)* 156.9 = -1.0983*10^-6 J