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

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(B) Let the distance between the two charges be $x$. The electric field $E$ at the position of charge $Q$ due to charge $-3Q$ is given by:$E = \frac{k

Vedclass · Accepted Answer

$E/3$

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(B) Let the distance between the two charges be $x$. The electric field $E$ at the position of charge $Q$ due to charge $-3Q$ is given by:$E = \frac{k| -3Q |}{x^2} = \frac{3kQ}{x^2}$Therefore,$\frac{kQ}{x^2} = \frac{E}{3}$.Now,the electric field $E'$ at the position of charge $-3Q$ due to charge $Q$ is given by:$E' = \frac{k| Q |}{x^2} = \frac{kQ}{x^2}$Substituting the value from the first equation:$E' = \frac{E}{3}$Since the electric field is a vector quantity and the field at $-3Q$ due to $Q$ points in the opposite direction to the field at $Q$ due to $-3Q$ (assuming the direction of $E$ is defined as positive),the magnitude is $E/3$.

Two charges $Q$ and $-3Q$ are placed at a certain distance $x$. If the electric field at the position of charge $Q$ due to charge $-3Q$ is $E$,then what is the electric field at the position of charge $-3Q$ due to charge $Q$?

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