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(D) Initially,$12$ positive charges of magnitude $q$ are placed symmetrically on a circle of radius $R$. Due to the symmetry of the arrangement,the el

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

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(D) Initially,$12$ positive charges of magnitude $q$ are placed symmetrically on a circle of radius $R$. Due to the symmetry of the arrangement,the electric force exerted by each charge on the central charge $Q$ is cancelled by the force exerted by the diametrically opposite charge. Thus,the net force on $Q$ is zero.When one charge $q$ is removed,the symmetry is broken. The forces from the remaining $11$ charges no longer cancel out. Specifically,the force that was previously exerted by the removed charge $q$ on $Q$ is now missing. Let the force that was exerted by the removed charge be $\vec{F}_{removed}$. Since the net force was zero,we have $\vec{F}_{net} + \vec{F}_{removed} = 0$,which implies $\vec{F}_{net} = -\vec{F}_{removed}$.The magnitude of the force exerted by a single charge $q$ on $Q$ at distance $R$ is $F = \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{R^2}$.Since the removed charge $q$ was positive and $Q$ is positive,the force $\vec{F}_{removed}$ was directed away from the position of the removed charge. Therefore,the net force on $Q$ after removing the charge is equal in magnitude to this force but directed towards the position of the removed charge. Thus,the force is $\frac{q Q}{4 \pi \varepsilon_0 R^2}$ 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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