A uniform electric field of strength E exists in a region. An electron (charge -e, mass m) enters a points A perpendicular to x-axis with velocity V . It moves through the electric field & exits at point B. Then:

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$E = - \frac{2 a m v^{2}}{e d^{2}} \hat{i}$

Rate of work done by the electric field at B is $\frac{4 m a^{2} v^{3}}{d^{3}}$

Rate of work by the electric field at A is zero

Velocity at B is $\frac{2 a v}{d} \hat{i} + v \hat{j}$

answer is A, B, C, D.

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Detailed Solution

As particle's velocity along Y axis ls not changing so it means electric field is not having any component in Y direction. As particle is having non-zero x-component of displacement in positive direction so it means electric field is along negative X-axis. From given information, $d = v \times t$ where t is the time taken by electron to move from A to B.

So, $a = \frac{1}{2} \times \frac{e E}{m} t^{2}$

Solving above equation we get, $E = - \frac{2 a m v^{2}}{e d} \hat{i}$

Rate of work done is given by $P = \vec{F} \cdot \vec{V} = F V_{x} = (e E) (\frac{e E}{m} . \frac{d}{v})$