Four point charges, each having a charge with a magnitude of 3.3 µC, are at the
ID: 1448550 • Letter: F
Question
Four point charges, each having a charge with a magnitude of 3.3 µC, are at the corners of a square whose sides are 4 m long. Find the electrostatic potential energy of this system under the following conditions.
(a) all of the charges are negative
(b) three of the charges are positive and one of the charges is negative
(c) the charges at two adjacent corners are positive and the other two charges are negative
(d) the charges at two opposite corners are positive and the other two charges are negative (Assume the potential energy is zero when the point charges are very far from each other.)
Explanation / Answer
PE of two charge system = kq1q2/d
A) for square there 6 sets of charges.
PE = [kq1q2 / d ] + [kq3q2 / d ] + [kq3q4 / d ] + [kq1q4 / d ] + [kq1q3 /sqrt(d^2 + d^2) ] + [kq4q2 / sqrt(d^2 + d^2) ]
PE = 4[ (9 x 10^9 x -3.3 x 10^-6 x -3.3 x 10^-6)/(4)] + 2[ (9 x 10^9 x -3.3 x 10^-6 x -3.3 x 10^-6)/sqrt(4^2 + 4^2)]
PE = 0.133 J
B)
PE = 4[ (9 x 10^9 x 3.3 x 10^-6 x 3.3 x 10^-6)/(4)] + 2[ (9 x 10^9 x 3.3 x 10^-6 x 3.3 x 10^-6)/sqrt(4^2 + 4^2)]
PE = 0.133 J
c) PE = [ (9 x 10^9 x 3.3 x 10^-6 x 3.3 x 10^-6)/(4)] + 2[ (9 x 10^9 x -3.3 x 10^-6 x 3.3 x 10^-6)/sqrt(4^2 + 4^2)]
+2[ (9 x 10^9 x -3.3 x 10^-6 x 3.3 x 10^-6)/(4)] + [ (9 x 10^9 x -3.3 x 10^-6 x -3.3 x 10^-6)/(4)]
PE = -0.035 J
d)
PE = [ (9 x 10^9 x 3.3 x 10^-6 x 3.3 x 10^-6)/(4)] + 2[ (9 x 10^9 x 3.3 x 10^-6 x 3.3 x 10^-6)/sqrt(4^2 + 4^2)]
+2[ (9 x 10^9 x -3.3 x 10^-6 x 3.3 x 10^-6)/(4)] + [ (9 x 10^9 x -3.3 x 10^-6 x -3.3 x 10^-6)/(4)]
PE = 0.035 J
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