Two point charges Q and -3Q are placed at a d | vedclass

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(D) Let the charges be at points $A$ and $B$ separated by distance $x$. The electric field at $A$ (location of $Q$) is due to the charge $-3Q$ at $B$.

Vedclass · Accepted Answer

$E/3$

Vedclass · Answer

(D) Let the charges be at points $A$ and $B$ separated by distance $x$. The electric field at $A$ (location of $Q$) is due to the charge $-3Q$ at $B$.$E = \frac{k| -3Q |}{x^2} = \frac{3kQ}{x^2}$.Now,the electric field at $B$ (location of $-3Q$) is due to the charge $Q$ at $A$.$E' = \frac{k|Q|}{x^2} = \frac{kQ}{x^2}$.Comparing the two expressions,we get $E' = \frac{E}{3}$.Since both fields are directed along the line $AB$ (the field at $A$ is towards $B$ and the field at $B$ is away from $A$),the magnitude is $E/3$.

Two point charges $Q$ and $-3Q$ are placed at a distance $x$ apart. If the electric field at the location of $Q$ is $E$,then the electric field at the location of $-3Q$ is:

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