Two fixed charges +4,Q and -Q are located at | vedclass

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(D) Let the point $P$ be at a distance $x$ from $B$ on the side of $B$. The electric field due to $+4\,Q$ at $A$ and $-Q$ at $B$ must be equal in magn

Vedclass · Accepted Answer

Both $A$ and $B$ are correct.

Vedclass · Answer

(D) Let the point $P$ be at a distance $x$ from $B$ on the side of $B$. The electric field due to $+4\,Q$ at $A$ and $-Q$ at $B$ must be equal in magnitude and opposite in direction.$E_A = E_B \implies \frac{k(4Q)}{(3+x)^2} = \frac{kQ}{x^2}$$4x^2 = (3+x)^2 \implies 2x = 3+x \implies x = 3\,m$.Thus,the point $P$ is $3\,m$ from $B$ outside $AB$. So,option $A$ is correct.Now consider the potential $V$ near $P$. The potential $V$ is given by $V = \frac{k(4Q)}{r_A} - \frac{kQ}{r_B}$.For a negative charge $q_n$ at $P$,the potential energy $U_n = q_n V$. Since $q_n$ is negative,the potential energy $U_n$ has a local minimum at $P$,implying stable equilibrium. Thus,a negative charge will oscillate if displaced slightly along the line $AB$. So,option $B$ is correct.For a positive charge $q_p$ at $P$,the potential energy $U_p = q_p V$ has a local maximum at $P$,implying unstable equilibrium. It will not oscillate. Thus,option $C$ is incorrect.Therefore,both $A$ and $B$ are correct.

Two fixed charges $+4\,Q$ and $-Q$ are located at $A$ and $B$ respectively,with the distance $AB = 3\,m$.

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