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$12$ 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
zero
$\frac{q Q}{4 \pi \varepsilon_0 R^2}$ away from the position of the removed charge
$\frac{11 q Q}{4 \pi \varepsilon_0 R^2}$ away from the position of the removed charge
$\frac{q Q}{4 \pi \varepsilon_0 R^2}$ towards the position of the removed charge
Solution

(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 $q\left(\right.$ let $\left.q_1\right)$ 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.