Question

A uniform electric field of magnitude $290V/m$ is directed in the positive $x$ direction. A $+13.0\mu C$ charge moves from the origin to the point $(x,y)=(20.0cm,50.0cm).$
What is the change in the potential energy of the charge field system?

• A. $-754J$
• B. $-754mJ$
• C. $-754kJ$
• D. $-754\mu J$

Answer 1

The correct option is D $-754\mu J$

Given electric field, $\overrightarrow{E}=290V/m$ directed along $+x$ direction.

Charge $=+13\mu C$ moves from origin to point $(x,y)=(20cm,50cm)$.

We have to find the charge in potential energy of the charge field system.

We know, the change in potential energy $=$ charge $\times$ change in potential

i.e, $\Delta U=q\Delta V$

But, $\Delta V=-Ed$

$\Delta U=-qEd$

$=-13\times {10}^{-6}\times 290\times 20\times {10}^{-2}$

$=0.000754J$

$=-754\times {10}^{-6}J$

Here $d=20cm$, since electric field is along positive $x-$axis.