Question:
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)
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)
(Take V = 0 at infinity)
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All charges are at the same distance R, so V is the algebraic sum of kq/R terms; E is the vector sum of five equal-magnitude vectors that are symmetric about the centre.
Updated On: Nov 5, 2026
- \(V = 0;\ E = 0\)
- \(V = \frac{10q}{4\pi\varepsilon_0 R};\ E = \frac{10q}{4\pi\varepsilon_0 R^2}\)
- \(V = \frac{10q}{4\pi\varepsilon_0 R};\ E = 0\)
- \(V = 0;\ E = \frac{10q}{4\pi\varepsilon_0 R^2}\)
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The Correct Option is A
Solution and Explanation
Potential at the centre (scalar sum): \(V = \frac{1}{4\pi\varepsilon_0 R}(5q - 5q) = 0\).
Electric field (vector sum): each charge \(q_i\) at angle \(\theta_i\) produces a field of magnitude \(\frac{|q|}{4\pi\varepsilon_0 R^2}\) along \(-q_i\,\hat r_i\) (away from +q, towards -q).
So \(\vec E = -\frac{1}{4\pi\varepsilon_0 R^2}\sum q_i \hat r_i = -\frac{q}{4\pi\varepsilon_0 R^2}\left[\sum_{j=0}^{4}\hat r(72^\circ j) - \sum_{j=0}^{4}\hat r(36^\circ + 72^\circ j)\right]\).
Five unit vectors equally spaced by \(72^\circ\) add to zero, and so do the other five. Hence \(\vec E = 0\).
Thus V = 0 and E = 0.
Answer: (1) V = 0; E = 0
Electric field (vector sum): each charge \(q_i\) at angle \(\theta_i\) produces a field of magnitude \(\frac{|q|}{4\pi\varepsilon_0 R^2}\) along \(-q_i\,\hat r_i\) (away from +q, towards -q).
So \(\vec E = -\frac{1}{4\pi\varepsilon_0 R^2}\sum q_i \hat r_i = -\frac{q}{4\pi\varepsilon_0 R^2}\left[\sum_{j=0}^{4}\hat r(72^\circ j) - \sum_{j=0}^{4}\hat r(36^\circ + 72^\circ j)\right]\).
Five unit vectors equally spaced by \(72^\circ\) add to zero, and so do the other five. Hence \(\vec E = 0\).
Thus V = 0 and E = 0.
Answer: (1) V = 0; E = 0
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