Q. An electric field is given by$\overrightarrow{E} = \left( y\widehat{\dot{i}} + x\widehat{\dot{j}} \right)m\ {NC}^{- 1}$. The work done in moving a 1 C charge from $\overrightarrow{r}$$_{A}$ =$(2\widehat{\dot{i}} + 2\widehat{\dot{j}}\ )$m to $\overrightarrow{r}$$_{B}$ =$(4\widehat{\dot{i}} + 2\widehat{\dot{j}}\ )$m is
A) $+ 8$ J
B) +4 J
C) Zero
D) $-$4 J
Explanation
(c) Work done$= Fs\cos\theta = F(2\pi r)\cos 90^{0} = 0$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.