South Florida Business Daily

How much work does it take to assemble this charge distribution?

Four 60-μC charges are brought from far apart onto a line where they're spaced at 1.4-cm intervals. How much work does it take to assemble this charge distribution?

Public Comments

1. You could assemble the charges one at a time but, since the E field forces add vectorially, the total work is the sum of works required to assemble all of the pairs. This is simplified by observing that all the charges have the same value q and the distances are integer multiples of the shortest distance r. Given q=60 uC and r = 1.4 cm, the work required to bring a pair of charges to a distance r apart is W1 = (q^2)*(4*pi*e0*r). The work required to bring a pair to a distance 2r apart is W2 = (W1)/2. (why?) And for a distance 3r, W3 = (W1)/3. In the final assembly, there are 3 pairs r apart, 2 pairs 2r apart and a single pair 3r apart. The total work is: W = 3*W1 + 2*W2 + W1 = (13/3)*W1