Question

A proton is moved 15 cm on path parallel to the field lines of a uniform electric field of $2.0\times {10}^5$ V/m
What are the possible change in potential ?consider both cases of a moving the proton .

Answer 1

We know, change in potential when a charge is moved in an Electric Field,

$\Delta =-E.\Delta x$ ( $\Delta x$: displacement )

$\left(i\right)$ when proton is displaced along $\overrightarrow{E}$ field, $\Delta x=15cm$

$\Delta v_1=-2.0\times {10}^5\times \left(0.15\right)$

$\Delta v_1=-30kv$

$\left(ii\right)$ when proton is displaced against $\overrightarrow{E}$ field, $\Delta x=-15cm$.

The potential difference is just negative of case $\left(i\right)$

$\Rightarrow \Delta v_2=-\Delta v_1=30kv$.