Question

An electric field is given by $\overrightarrow{E}=(y\hat{i}+\hat{x})N{C}^{-1}$. Find the work done (in $J$) by the electric field in moving a $1C$ charge from ${\overrightarrow{r}}_A=(2\hat{i}+2j)m$ to ${\overrightarrow{r}}_B=(4\hat{i}+\hat{j})m$

• A. $0J$
• B. $-2J$
• C. $2J$
• D. $4J$

Answer 1

The correct option is A $0J$

Work done , $W=\int \overrightarrow{F}.\overrightarrow{dr}$

Here electrostatic force , $\overrightarrow{F}=q\overrightarrow{E}=q(y\hat{i}+x\hat{j})$
$\overrightarrow{F}.\overrightarrow{dr}=q(y\hat{i}+x\hat{j}).(dx\hat{i}+dy\hat{j})=q(ydx+xdy)=d\left(xy\right)$ as $q=1C$

Now $W={\int }_{2,2}^{4,1}d\left(xy\right)=[xy{]}_{2,2}^{4,1}=4\times 1-2\times 2=4-4=0$