Question

An electric field of $20N{C}^{-1}$ exists along the x-axis in space. Calculate the potential difference $V_B-V_A$ where the points A and B are given by, $A=(0,0);B=(4m,2m)$

A charge of $-2.0\times {10}^{-4}$C is moved from the point A to the point B. Find the change in electrical potential energy $U_B-U_A$.

Answer 1

The magnitude of the charge is $-2.0\times {10}^{-4}C$

The Electric field is along x-direction

Thus potential difference between $(0,0)$ and $(4,2)$ is,

Here, $\Delta x=4m$.

$\Delta V=-E\times \Delta x$

$=-20\times \left(4\right)=-80V$

Therefore, the potential energy $(U_B-U_A)$ between the points

$U=\Delta V\times q$

$=-80\times (-2)\times {10}^{-4}=160\times {10}^{-4}=0.016J$.