The point charges Q and -2Q are placed at a d | vedclass

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(C) Let the position of charge $Q$ be $A$ and the position of charge $-2Q$ be $B$. The distance between them is $r$.The electric field at point $A$ du

Vedclass · Accepted Answer

$+\frac{\vec E}{2}$

Vedclass · Answer

(C) Let the position of charge $Q$ be $A$ and the position of charge $-2Q$ be $B$. The distance between them is $r$.The electric field at point $A$ due to charge $-2Q$ at $B$ is given by:$\vec E = \frac{k(-2Q)}{r^2} \hat{r}_{BA} = \frac{2kQ}{r^2} \hat{r}_{AB}$ (where $\hat{r}_{AB}$ is the unit vector from $A$ to $B$).The electric field at point $B$ due to charge $Q$ at $A$ is given by:$\vec E' = \frac{kQ}{r^2} \hat{r}_{AB}$.Comparing the two expressions:$\vec E' = \frac{1}{2} \left( \frac{2kQ}{r^2} \hat{r}_{AB} \right) = \frac{\vec E}{2}$.Since both fields point in the same direction (from $A$ to $B$),the electric field at $B$ is $+\frac{\vec E}{2}$.

The point charges $Q$ and $-2Q$ are placed at a distance $r$ apart. If the electric field at the location of $Q$ (point $A$) due to $-2Q$ is $\vec E$,then the electric field at the location of $-2Q$ (point $B$) due to $Q$ will be:

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