Physics
posted by Anne .
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 = 7.00 109 C placed at corner A and then at corner B.

diagonal of the square = .25 * √(2)
√(2) is the square root of 2 (which equals 1.414)
corner A:

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:

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