Question

A uniform electric field of strength $E$ exists in a region. An electron enters a point $A$ with velocity $v$ as shown. It moves through the field and reaches a point $B$. Velocity of particle at $B$ is $2v$ at ${30}^o$ with $x-$axis. Then,

• A. electric field $E=-\frac{3m{v}^2}{2ea}\hat{i}$
• B. rate of doing work done by electric field at $B$ is $\frac{3m{v}^3}{2ea}$
• C. Both $\left(a\right)$ and $\left(b\right)$ are correct
• D. Both $\left(a\right)$ and $\left(b\right)$ are wrong

Answer 1

The correct option is A electric field $E=-\frac{3m{v}^2}{2ea}\hat{i}$

Using work-energy theorem:
$W_{all}=\Delta K.E$
$\overrightarrow{F}.\overrightarrow{d}=\Delta K.E$
Since only the velocity in x-axis is changing, this means force is only in x-axis(so take displacement in x-axis only). Also since charge under consideration is electron(-ve), force will be opposite of the direction of E.
$\left(eE\right)a=\frac{1}{2}m(4v^2-v^2)$
$\overrightarrow{E}=-\frac{3m{v}^2}{2e}\hat{i}$