SOLUTION

The electrostatic field is given by

$E = \left( {\dfrac{x}{2} + 2y} \right)\widehat i + 2x\widehat j$

The work is done in moving a point charge Q = −20 μC from the origin (0, 0, 0) to point (4, 0, 0)

W = V × Q

The  potential difference from point A to B is given by

$V = – \int\limits_A^B {\overrightarrow E .\overrightarrow {dl} }$

As y ordinate doesn’t change along the path, therefore, y = 0, then dy = 0

$V = – \int\limits_0^4 {\dfrac{1}{2}xdx = – \dfrac{1}{4}\left[ {{x^2}} \right]_0^4}$

V = −4 V

W = V × Q

W = −4 × −20

W = 80 μJ