Question

An electron is taken from point $A$ to point $B$ along the path $AB$ in a uniform electric field of intensity $E=10\text{V/m}$. Side $AB=5\text{m}$, and side $BC=3\text{m}$. Then, the amount of external workdone on the electron (in $\text{eV})$ is

Answer 1

From the given figure, we can deduce that, $B$ and $C$ are on the same equipotential surface. Thus, potential at $B$ is same as potential at $C$.

Given,

Length $AB=5\text{m}$
Length $BC=3\text{m}$
$\therefore AC=\sqrt{{\left(AB\right)}^2-{\left(BC\right)}^2}$

$\Rightarrow AC=\sqrt{{\left(5\right)}^2-{\left(3\right)}^2}=4\text{m}$

Then,

Potential difference ,$V_C-V_A=-E\times AC$

$\Rightarrow V_C-V_A=-10\times 4=-40\text{volts}$

Since, Workdone $W=q\times (V_C-V_A)$

we get,

$W=-e\times (V_C-V_A)$

Substituting the data obtained,

$W=-e\times (-40)=40eV$

Accepted answer : $40$