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

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(B) Let the charges be at points $A$ and $B$ separated by a distance $r$. Charge at $A$ is $Q$ and charge at $B$ is $-3 Q$.The electric field at $A$ d

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$(E / 3) \hat{i}$

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(B) Let the charges be at points $A$ and $B$ separated by a distance $r$. Charge at $A$ is $Q$ and charge at $B$ is $-3 Q$.The electric field at $A$ due to charge at $B$ is given as $\vec{E}_A = E \hat{i}$.The magnitude of the electric field at $A$ due to $-3 Q$ is $E = \frac{k | -3Q |}{r^2} = \frac{3kQ}{r^2}$.Thus,$\frac{kQ}{r^2} = \frac{E}{3}$.Now,the electric field at $B$ due to charge at $A$ $(Q)$ is directed away from $A$ (since $Q$ is positive). If we assume the vector from $A$ to $B$ is along $\hat{i}$,then the field at $B$ due to $A$ is $\vec{E}_B = \frac{kQ}{r^2} \hat{i}$.Substituting the value of $\frac{kQ}{r^2}$,we get $\vec{E}_B = \frac{E}{3} \hat{i}$.

Two point charges $Q$ and $-3 Q$ are placed at some distance apart. If the electric field due to $-3 Q$ at the location of $Q$ is $E \hat{i}$,then the electric field due to $Q$ at the location of $-3 Q$ is:

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