Question

$12$ positive charges of magnitude $q$ are placed on a circle of radius $R$ in a manner that they are equally spaced. A charge $Q$ is placed at the centre, if one of the charges $q$ is removed, then the force on $Q$ is

• A. zero
• B. $\frac{q Q}{4 \pi \varepsilon_{0} R^{2}}$ away from the position of the removed charge
• C. $\frac{11 q Q}{4 \pi \varepsilon_{0} R^{2}}$ away from the position of the removed charge
• D. $\frac{q Q}{4 \pi \varepsilon_{0} R^{2}}$ towards the position of the removed charge

Answer 1

Answer: (D)

Force on charge $Q$ is initially zero as forces of $12$ charges balances each other.

As shown in above figure, forces of diametrically opposite charges balances each other, hence net force on $Q$ is zero. When one of the charge $q$ (let $q_{1}$ ) is removed, net force on $Q$ is now the unbalanced force of diametrically opposite charge.
i.e. Force, $F=\frac{k q Q}{R^{2}}=\frac{q Q}{4 \pi \varepsilon_{0} R^{2}}$
and this force vector points towards the position of the removed charge.