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(D) Let the electric field due to a single charge $q$ at the centre $O$ be $\vec{E}_0$. In a regular hexagon,the electric fields due to charges at opp

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(D) Let the electric field due to a single charge $q$ at the centre $O$ be $\vec{E}_0$. In a regular hexagon,the electric fields due to charges at opposite corners cancel each other out. With five charges $q$ placed as shown,the electric field due to the four charges at the corners (forming two pairs of opposite charges) cancels out,leaving only the field due to the single charge $q$ at the rightmost corner. Thus,the net electric field at $O$ is $\vec{E} = \vec{E}_0$ directed towards the left. To get a net electric field of $6\vec{E}$ at $O$,we need to place a charge $Q'$ at the empty sixth corner such that the total field becomes $6\vec{E}_0$. The field due to the existing five charges is $\vec{E}_0$. Let the field due to the new charge $Q'$ be $\vec{E}'$. Then $\vec{E}_0 + \vec{E}' = 6\vec{E}_0$,which implies $\vec{E}' = 5\vec{E}_0$. Since $\vec{E}_0$ is directed towards the left,$\vec{E}'$ must also be directed towards the left. $A$ charge $Q'$ placed at the left corner produces a field directed towards the right. To produce a field directed towards the left,we must place a negative charge. Specifically,a charge of $-5q$ at the left corner will produce an electric field of $5\vec{E}_0$ directed towards the left. Therefore,the required charge is $-5q$.

Five point charges,each having magnitude $q$,are placed at the corners of a hexagon as shown in the figure. The net electric field at the centre $O$ is $\vec{E}$. To make the net electric field at $O$ equal to $6\vec{E}$,the charge to be placed on the remaining sixth corner should be: (in $q$)

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