Question

An electric field is given by $\overrightarrow{E}=(y\hat{i}+x\hat{j})\text{N/C}$. The work done in moving a $1\text{C}$ charge from $\overrightarrow{r_A}=(2\hat{i}+2\hat{j})\text{m}$ to $\overrightarrow{r_B}=(4\hat{i}+\hat{j})\text{m}$ is

• A. $+4\text{J}$
• B. $-4\text{J}$
• C. $+8\text{J}$
• D. Zero

Answer 1

The correct option is D Zero

$A=(2,2)$ and $B=(4,1)$
Now $W_{A\to B}=q(V_B-V_A)\dots \dots \left(1\right)$
$\underset{A}{\overset{B}{\int }}dV=-\underset{A}{\overset{B}{\int }}\overrightarrow{E}.d\overrightarrow{r}$
$\Rightarrow V_B-V_A=-\underset{(2,2)}{\overset{(4,1)}{\int }}(y\hat{i}+x\hat{j}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$
$\Rightarrow V_B-V_A=-\underset{(2,2)}{\overset{(4,1)}{\int }}(ydx+xdy)$
$=-\underset{(2,2)}{\overset{(4,1)}{\int }}d\left(xy\right)=[-xy{]}_{2,2}^{4,1}=0$
$\therefore W_{A\to B}=0$ [from Eq. (1)]