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(C) Let the distance from the center $O$ to any vertex be $a$. The electric field at $O$ due to a charge $q$ at any vertex is $E = \frac{kq}{a^2}$ dir

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the distance from the center $O$ to any vertex be $a$. The electric field at $O$ due to a charge $q$ at any vertex is $E = \frac{kq}{a^2}$ directed away from the charge.For a regular hexagon,the electric field at the center due to charges at opposite vertices $P$ and $S$,$Q$ and $T$,$R$ and $U$ can be calculated.Let the charges at $P, Q, R, S, T, U$ be $q_1, q_2, q_3, q_4, q_5, q_6$ respectively.The net electric field at $O$ is $\vec{E}_{net} = \sum_{i=1}^6 \vec{E}_i$.Given that the net field is twice the field due to a single charge $+q$ at $R$,i.e.,$E_{net} = 2 \left( \frac{kq}{a^2} \right)$.If we place $-q$ at $U$ and $+q$ at $R$,the field at $O$ due to these two is $\vec{E}_R + \vec{E}_U = \frac{kq}{a^2} \hat{r} + \frac{kq}{a^2} \hat{r} = 2 \frac{kq}{a^2} \hat{r}$.To keep the net field as $2 \frac{kq}{a^2}$,the remaining four charges at $P, Q, S, T$ must cancel each other out.This happens if we place equal and opposite charges at opposite vertices,e.g.,$q_P = -q_S$ and $q_Q = -q_T$.Checking the options,option $C$ $(-, +, +, -, +, -)$ corresponds to $P=-q, Q=+q, R=+q, S=-q, T=+q, U=-q$.Here,$P$ and $S$ are opposite and have charges $-q$ and $-q$ (not cancelling),but let's re-evaluate: The field at $O$ due to $P$ and $S$ is $\vec{E}_P + \vec{E}_S$. If $P=-q$ and $S=-q$,they don't cancel. Actually,for option $C$: $P=-q, Q=+q, R=+q, S=-q, T=+q, U=-q$. $P$ and $S$ are opposite: $-q$ and $-q$ (Field = $0$).$Q$ and $T$ are opposite: $+q$ and $+q$ (Field = $0$).$R$ and $U$ are opposite: $+q$ and $-q$ (Field = $2E$ towards $U$).Thus,the total field is $2E$ towards $U$. This matches the condition.

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