Six charges,three positive and three negative | vedclass

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(D) Let the magnitude of each charge be $q$ and the distance from the center $O$ to any corner be $a$. The electric field due to a charge $q$ at a cor

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$-, +, +, -, +, -$

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(D) Let the magnitude of each charge be $q$ and the distance from the center $O$ to any corner be $a$. The electric field due to a charge $q$ at a corner is $E = \frac{kq}{a^2}$.If we place charges at $P, Q, R, S, T, U$ as $(-, +, +, -, +, -)$,the fields at $O$ are:- Field due to $P$ (at $P$) and $S$ (at $S$): $P$ has $-q$ and $S$ has $-q$. The field due to $P$ is $\vec{E}_P$ away from $P$ (towards $S$) and field due to $S$ is $\vec{E}_S$ away from $S$ (towards $P$). Since they are equal and opposite,they cancel out.- Similarly,$Q$ $(+q)$ and $T$ $(-q)$ produce fields that do not cancel. $Q$ produces field $\vec{E}_Q$ away from $Q$ (towards $T$) and $T$ produces field $\vec{E}_T$ towards $T$ (away from $T$). These add up to $2E$ in the direction from $Q$ to $T$.- $R$ $(+q)$ and $U$ $(-q)$ produce fields $\vec{E}_R$ away from $R$ (towards $U$) and $\vec{E}_U$ towards $U$ (away from $U$). These add up to $2E$ in the direction from $R$ to $U$.- The resultant field is the vector sum of these components. For the configuration $(-, +, +, -, +, -)$,the net field at $O$ is $2E$ in the direction of the vector sum of the components,which satisfies the condition of being double the field of a single charge at $R$.

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