Question

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

• A. $+4J$
• B. $-4J$
• C. $+8J$
• D. Zero

Answer 1

Answer: (D)

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