An electrostatic field is given by $E = \left( {\dfrac{x}{2} + 2y} \right)\widehat i + 2x\widehat j$ V / m .
An electrostatic field is given by $E = \left( {\dfrac{x}{2} + 2y} \right)\widehat i + 2x\widehat j$ V / m . Find the work done in moving a point charge Q = −20 μC from the origin to (4,0,0) m. $(\widehat i,\widehat j)$ are the unit vectors along x,y-axis.
Right Answer is:
80 J
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