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Two positive charges of magnitude q are placed at the two ends of one side (call it side 1) of a square of side 2a. Two negative charges of the same magnitude q are placed at the other two corners. A charge Q starts from rest at the midpoint of side 1 and moves to the centre of the square. Its kinetic energy at the centre is:

1. (1/(4*pi*epsilon₀)) * (2qQ/a) * (1 - 1/sqrt(5))
2. zero
3. (1/(4*pi*epsilon₀)) * (2qQ/a) * (1 + 1/sqrt(5))
4. (1/(4*pi*epsilon₀)) * (2qQ/a) * (1 - 2/sqrt(5))

Correct answer: (1/(4*pi*epsilon₀)) * (2qQ/a) * (1 - 1/sqrt(5))

Solution

Place the square with side 2a. The two +q charges are at the ends of side 1; the two -q charges at the far corners. At the centre, every charge is distance a*sqrt(2) away and the +q and -q contributions cancel, so V_centre = 0. At the midpoint of side 1: distance to each +q is a (so V from both +q = 2*kq/a), distance to each -q is sqrt((2a)² + a²) = a*sqrt(5) (so V from both -q = -2*kq/(a*sqrt(5))). Hence V_mid = (2kq/a)(1 - 1/sqrt(5)). KE at centre = Q*(V_mid - V_centre) = Q*(2kq/a)(1 - 1/sqrt(5)) = (1/(4*pi*epsilon₀))*(2qQ/a)*(1 - 1/sqrt(5)).

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