Five point charges each having magnitude q ar | vedclass

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(C) Let the side length of the regular hexagon be $a$. The electric field at the centre $O$ due to a charge $q$ at any corner is $E_0 = \frac{kq}{a^2}

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$5$

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(C) Let the side length of the regular hexagon be $a$. The electric field at the centre $O$ due to a charge $q$ at any corner is $E_0 = \frac{kq}{a^2}$ directed away from the charge.In a regular hexagon,the electric fields due to charges at opposite corners cancel each other out.With five charges $q$ at five corners,the fields due to four charges (two pairs of opposite corners) cancel out,leaving the field due to the single charge at the remaining corner. Let this field be $\vec E = \frac{kq}{a^2}$ directed away from the empty corner.To get a net electric field of $6\vec E$ at the centre,we need a field of $6\vec E$ directed away from the sixth corner.Let the charge at the sixth corner be $Q'$. The field due to this charge at the centre is $\vec E' = \frac{kQ'}{a^2}$.The total field at the centre is $\vec E_{net} = \vec E + \vec E' = \vec E + \frac{kQ'}{a^2} = 6\vec E$.This implies $\frac{kQ'}{a^2} = 5\vec E = 5\left(\frac{kq}{a^2}\right)$.Therefore,$Q' = 5q$.

Five point charges each having magnitude $q$ are placed at the corners of a regular 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 placed on the remaining sixth corner should be: (in $,q$)

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