Two charges +Q and -2 Q are located at points | vedclass

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(B) Let the distance between charges $A$ and $B$ be $d$. Let the point where the electric field is zero be at a distance $x$ from charge $A$ on the li

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left of $A$ on the line

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(B) Let the distance between charges $A$ and $B$ be $d$. Let the point where the electric field is zero be at a distance $x$ from charge $A$ on the line joining them.For the electric field to be zero,the magnitudes of the electric fields due to both charges must be equal,and their directions must be opposite.$1$. Between $A$ and $B$: The fields due to $+Q$ and $-2 Q$ are in the same direction,so the net field cannot be zero.$2$. Right of $B$: The field due to $-2 Q$ is always stronger than the field due to $+Q$ because $-2 Q$ is larger in magnitude and closer to any point in this region. Thus,the net field cannot be zero.$3$. Left of $A$: Let the point be at a distance $x$ from $A$. The electric field due to $+Q$ is $E_A = \frac{kQ}{x^2}$ (directed to the left) and the field due to $-2 Q$ is $E_B = \frac{k(2Q)}{(d+x)^2}$ (directed to the right).Setting $E_A = E_B$,we get $\frac{kQ}{x^2} = \frac{2kQ}{(d+x)^2}$,which simplifies to $(d+x)^2 = 2x^2$,or $d+x = \sqrt{2}x$. This gives $x = \frac{d}{\sqrt{2}-1}$,which is a finite distance.Therefore,the electric field is zero at a point to the left of $A$.

Two charges $+Q$ and $-2 Q$ are located at points $A$ and $B$ on a horizontal line as shown below. The electric field is zero at a point which is located at a finite distance:

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