Question

A uniform electric field of strength E exists in a region. An electron (charge = –e and mass m) enters a point A(a, 0) with velocity v $\hat{j}$. It moves through the electric field and exits at point B(2a, d) with velocity $u\hat{i}+v\hat{j}$. Then:

• A. $\overrightarrow{E}=-\frac{2am{v}^2}{e{d}^2}\hat{i}$
• B. $u=\frac{2av}{d}$
• C. Rate of work done by the electric field at A is zero.
• D. Rate of work done by the electric field at B is $\frac{4m{a}^2{v}^3}{d^3}.$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\overrightarrow{E}=-\frac{2am{v}^2}{e{d}^2}\hat{i}$
B $u=\frac{2av}{d}$
C Rate of work done by the electric field at A is zero.
D Rate of work done by the electric field at B is $\frac{4m{a}^2{v}^3}{d^3}.$

For x - axis
$a=\frac{1}{2}\left(\frac{eE}{m}\right)t^2......\left(i\right)$
and for y - axis
d = vt...................(ii)
Solve (i) and (ii)
$E=\frac{2am{v}^2}{e{d}^2}and\overrightarrow{E}=\frac{-2am{v}^2}{e{d}^2}\hat{i}u=0+\left(\frac{eE}{m}\right).\frac{d}{v}$
Rate of W.D. = $\overrightarrow{F}.\overrightarrow{v}=\frac{eE}{m}.v$
at B, rate of W.D. = $\frac{4m{a}^2{v}^3}{d^3}$
at A, rate of W.D. = 0