Question

Four charges are arranged at the corners of a square ABCD as shown in figure. The force on a positive charge 'Q' kept at the centre of the square is :

• A. zero
• B. along diagonal AC
• C. along diagonal BD
• D. perpendicular to the side AB

Answer 1

The correct option is D perpendicular to the side AB

The force on $+Q$ at the center of square is,

$\overrightarrow{F}=k[\frac{-qQ(\overrightarrow{r_0}-\overrightarrow{r_1})}{|\overrightarrow{r_0}-\overrightarrow{r_1}{|}^3}+\frac{qQ(\overrightarrow{r_0}-\overrightarrow{r_2})}{|\overrightarrow{r_0}-\overrightarrow{r_2}{|}^3}+\frac{-2qQ(\overrightarrow{r_0}-\overrightarrow{r_3})}{|\overrightarrow{r_0}-\overrightarrow{r_3}{|}^3}+\frac{2qQ(\overrightarrow{r_0}-\overrightarrow{r_4})}{|\overrightarrow{r_0}-\overrightarrow{r_4}{|}^3}]$

$=\frac{kqQ}{r^2}\left[\right(\hat{i}+\hat{j})-(-\hat{i}+\hat{j})+2(-\hat{i}-\hat{j})-2(\hat{i}-\hat{j}\left)\right]$

$=\frac{kqQ}{r^2}[\hat{i}+\hat{j}+\hat{i}-\hat{j}-2\hat{i}-2\hat{j}-2\hat{i}+2\hat{j}]$

$=\frac{kqQ}{r^2}(-2\hat{i})$

thus, $F$ is perpendicular to side AB.