Question

Positive point charges of magnitude are placed at all the twelve 'hour' positions of a clock of radius $r$. The clock is mounted on a wall in the normal way. The charge at the position ${}^{\prime }{6}^{\prime }$ is removed The resulting electric field at the centre of the clock is___?

• A. $0$
• B. $\frac{1}{4\pi {ϵ}_o}\frac{q}{r^2}$ in the horizontal direction
• C. $\frac{1}{4\pi {ϵ}_o}\frac{q}{r^2}$ vertically upward
• D. $\frac{1}{4\pi {ϵ}_o}\frac{q}{r^2}$ vertically downward

Answer 1

Answer: (D)

$\frac{1}{4\pi {ϵ}_o}\frac{q}{r^2}$ vertically downward

the resultant force at centre is zero. On removing the charge from position 6, the resultant force at the centre will be $\frac{kq}{r^2}$ downward.