Question

A uniform electric field of magnitude $250V/m$ is directed in the positive x-direction. A + $12$ $\mu$C charge moves from the origin to the point $(x,y)=(20.0cm,5.0cm)$. What was the change in the potential energy of this charge?

• A. $+6\times {10}^{-4}J$
• B. $-6\times {10}^{-4}J$
• C. $-3\times {10}^{-4}J$
• D. $+3\times {10}^{-4}J$

Answer 1

Answer: (B)

$-6\times {10}^{-4}J$

Here the x-component of electric field is given only, i.e $E_x=250V/m$

Thus, $E_x=-\frac{dV}{dx}$

or $V=-{\int }_0^{0.2}{E}_{x}dx$ (as $x$ varies from $x=0$ to $x=20cm=0.2m$)

or $V=-250\left(0.2\right)=-50V$

The change in potential energy $=qV=12\times {10}^{-6}\times (-50)=-6\times {10}^{-4}J$