JEE PYQ: Electrostatics Question 25
Question 25 - 2020 (05 Sep Shift 2)
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) $V = \frac{10q}{4\pi\varepsilon_0 R};; E = 0$
(2) $V = 0;; E = \frac{10q}{4\pi\varepsilon_0 R^2}$
(3) $V = 0;; E = 0$
(4) $V = \frac{10q}{4\pi\varepsilon_0 R};; E = \frac{10q}{4\pi\varepsilon_0 R^2}$
Show Answer
Answer: (3)
Solution
Total charge $= 5q + 5(-q) = 0$, so $V = 0$. By symmetry (equal charges at equal angles, 72 degrees apart), the electric field vectors cancel out: $E = 0$.
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