12 positive charges of magnitude q are placed | vedclass

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Question

(d)

Force on charge $Q$ is initially zero as forces of $12$ charges balances each other.

As shown in above figure, forces of diametrically oppos

Vedclass · Accepted Answer

$\frac{q Q}{4 \pi \varepsilon_0 R^2}$ towards the position of the removed charge

Vedclass · 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 $q\left(\right.$ 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.

$12$ 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

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