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 vertex be $a$. The electric field at $O$ due to a charge $q$ a

Vedclass · Accepted Answer

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

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(D) Let the magnitude of each charge be $q$ and the distance from the center $O$ to any vertex be $a$. The electric field at $O$ due to a charge $q$ at a vertex is $E = \frac{kq}{a^2}$.If only one positive charge $q$ is at $R$,the electric field at $O$ is $E_0 = E$ (directed away from $R$).We want the resultant electric field at $O$ to be $2E$.For the arrangement in option $(d)$,the charges are: $P(-), Q(+), R(+), S(-), T(+), U(-)$.The electric field at $O$ due to $P$ and $S$ (both negative) cancel each other because they are opposite. Similarly,the electric field due to $Q$ and $T$ (both positive) cancel each other.This leaves the charges at $R$ and $U$. $R$ has a positive charge $+q$,creating a field $E$ directed away from $R$. $U$ has a negative charge $-q$,creating a field $E$ directed towards $U$. Since $R$ and $U$ are diametrically opposite,the field from $U$ (towards $U$) is in the same direction as the field from $R$ (away from $R$).Thus,the total electric field is $E + E = 2E$.

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