Explanation

(b) Here, mass of particle, $m = 2 \times 10^{- 5}Kg$
Charge of a particle, $q = 4 \times 10^{- 3}C$
Electric field, $E = 5\ V\ m^{- 1}$
Force on a charged particle in a uniform electric field is
$F = qE$
$\therefore$ Acceleration of the particle, $a = \frac{qE}{m}$
$= \frac{4 \times 10^{- 3} \times 5}{2 \times 10^{- 5}} = 10^{3}ms^{- 2}$
Let $v$ be velocity of particle after $10\ s$
As $v = u + at$
$\therefore v = at = \left( 10^{3}ms^{- 2} \right)(10\ s) = 10^{4}ms^{- 1}\ \ \ \lbrack u = 0\rbrack$
Kinetic energy of a particle after 10 s is
$K = \frac{1}{2}mv^{2} = \frac{1}{2} \times 2 \times 10^{- 5} \times 10^{4} \times 10^{4} = 10^{3}\ J$