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A charge Q is uniformly distributed along a thin rod AB of length L. Point O lies on the line of the rod at a distance L from the nearer end A. Find the electric potential at O.

1. Q / (8*pi*epsilon₀*L)
2. 3Q / (4*pi*epsilon₀*L)
3. Q / (4*pi*epsilon₀*L*ln 2)
4. Q*ln 2 / (4*pi*epsilon₀*L)

Correct answer: Q*ln 2 / (4*pi*epsilon₀*L)

Solution

Linear charge density lambda = Q/L. Take a small element dx at distance x from O; the rod spans x = L (end A) to x = 2L (end B). dV = k*lambda*dx/x. Integrating: V = k*lambda*ln(2L/L) = k*lambda*ln 2 = (1/(4*pi*epsilon₀))*(Q/L)*ln 2 = Q*ln2/(4*pi*epsilon₀*L).

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