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Two charges, 4q and -q, are fixed on the x-axis at x = -d/2 and x = +d/2 respectively. A third charge q is carried from the origin to the point x = d along a semicircular path (as shown). What is the change in the potential energy of charge q?

1. increase by 2*q²/(3*pi*eps0*d)
2. increase by 3*q²/(4*pi*eps0*d)
3. decrease by 4*q²/(3*pi*eps0*d)
4. decrease by q²/(4*pi*eps0*d)

Correct answer: decrease by 4*q²/(3*pi*eps0*d)

Solution

Let k = 1/(4*pi*eps0). At the origin: distance to 4q (at -d/2) is d/2, to -q (at d/2) is d/2. V_i = k(4q/(d/2) + (-q)/(d/2)) = k*(8q - 2q)/d = 6kq/d. At x = d: distance to 4q is 3d/2, to -q is d/2. V_f = k(4q/(3d/2) + (-q)/(d/2)) = k(8q/(3d) - 2q/d) = k(8q/3 - 6q/3)/d = k*(2q/3)/d = 2kq/(3d). Change in U = q(V_f - V_i) = kq(2q/(3d) - 6q/d) = kq²(2/3 - 6)/d = kq²(-16/3)/d. With k = 1/(4*pi*eps0): dU = -16q²/(3*4*pi*eps0*d) = -4q²/(3*pi*eps0*d). Negative -> energy decreases by 4q²/(3*pi*eps0*d).

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