Question

An electric field is given by $\vec{E}=(y \hat{i}+x \hat{j}) N C^{-1}$. The work done in moving a $1 C$ 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)

$A=\left(\right.2,2\left.\right)$ and $B=\left(\right.4,1\left.\right)$
Now $W_{A \rightarrow B}=q\left(V_{B} - V_{A}\right)$ ... (1)
$\int\limits _{A}^{B}dV=-\int\limits _{A}^{B}\overset{ \rightarrow }{E}\cdot \overset{ \rightarrow }{d}r$
or $V_{B}-V_{A}=-\int\limits _{\left(2,2\right)}^{\left(4,1\right)}\left(y \hat{i} + x \hat{j}\right)\cdot \left(d x \hat{i} + d y \hat{j} + d z \hat{k}\right)$
or $V_{B}-V_{A}=-\int\limits _{\left(2,2\right)}^{\left(4,1\right)}\left(y d x + x d y\right)$
$=-\int\limits _{\left(2,2\right)}^{\left(4,1\right)}d\left(x y\right)=\left(\left[- x y\right]\right)_{2 ,2}^{4 ,1}=0$
$\therefore $ $W_{A \rightarrow B}=0$ [from Eq. (1)]