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Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (−q) each. The potential V and the electric field E at the centre of the circle are respectively. (Take V = 0 at infinity)

1. V = 10q / (4π ε₀ R); E = 10q / (4π ε₀ R²)
2. V = 0, E = 10q / (4π ε₀ R²)
3. V = 0, E = 0
4. V = 10q / (4π ε₀ R); E = 0

Correct answer: V = 0, E = 0

Solution

All 10 charges are equidistant R from the centre, so V = (k/R)(5(+q)+5(-q)) = 0. For E, summing the field vectors: the cosine/sine sums over the five +q positions (0,72,144,216,288 deg) and over the five -q positions (36,108,180,252,324 deg) each vanish (sum of 5 equally-spaced unit vectors = 0), so E_x = E_y = 0. Hence V = 0 and E = 0.

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