The point charges +q, -q, -q, +q, +Q and -q a | vedclass

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Question

(A) Let the distance from the centre $O$ to each vertex be $r$. The electric field due to a charge $q$ at distance $r$ is $E = \frac{kq}{r^2}$.Let $\v

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$\frac{q}{2}$

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(A) Let the distance from the centre $O$ to each vertex be $r$. The electric field due to a charge $q$ at distance $r$ is $E = \frac{kq}{r^2}$.Let $\vec{E}_A, \vec{E}_B, \vec{E}_C, \vec{E}_D, \vec{E}_E, \vec{E}_F$ be the electric fields at $O$ due to charges at $A, B, C, D, E, F$ respectively.The charges are: $A(+q), B(-q), C(-q), D(+q), E(+Q), F(-q)$.Electric fields at $O$:$\vec{E}_A$ is directed away from $A$ (towards $D$).$\vec{E}_D$ is directed away from $D$ (towards $A$).Since $q_A = q_D = +q$,$\vec{E}_A + \vec{E}_D = 0$.Similarly,$\vec{E}_B$ is directed towards $B$ (away from $E$),and $\vec{E}_E$ is directed away from $E$ (towards $B$).$\vec{E}_C$ is directed towards $C$ (away from $F$),and $\vec{E}_F$ is directed towards $F$ (away from $C$).Let $\vec{E}_0$ be the resultant field due to charges at $A, B, C, D, F$. $\vec{E}_0 = \vec{E}_A + \vec{E}_B + \vec{E}_C + \vec{E}_D + \vec{E}_F$.Since $\vec{E}_A + \vec{E}_D = 0$,we have $\vec{E}_0 = \vec{E}_B + \vec{E}_C + \vec{E}_F$.All these are directed towards the respective vertices (as they are negative charges $-q$): $\vec{E}_B$ towards $B$,$\vec{E}_C$ towards $C$,$\vec{E}_F$ towards $F$.By symmetry,the resultant of these three fields is a vector of magnitude $\frac{kq}{r^2}$ directed towards $E$.Given: $|\vec{E}_0| = 2 |\vec{E}_E|$,where $\vec{E}_E$ is the field due to $+Q$ at $E$.$\frac{kq}{r^2} = 2 \frac{kQ}{r^2} \implies Q = \frac{q}{2}$.

Similar Questions

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