Question

A charged particle placed in an electric field falls from rest through a distance $d$ in time $t$. If the charge on the particle is doubled, the time of fall through the same distance will be:

• A. $2t$
• B. $T$
• C. $\frac{t}{\sqrt{2}}$
• D. $\frac{t}{2}$

Answer 1

The correct option is C $\frac{t}{\sqrt{2}}$

Initial velocity $u=0$
Force on charge$=F=qE$
$F=ma$
$qE=ma$
$\frac{qE}{m}=a$
$s=ut+\frac{1}{2}a{t}^2$
$d=\frac{1}{2}\frac{qE}{m}{t}_1^2....\left(1\right)$
$d=\frac{1}{2}\frac{qE}{m}{t}_2^2....\left(2\right)$
$\frac{1}{2}\frac{qE}{m}{t}_1^2=\frac{1}{2}\frac{qE}{m}{t}_2^2\Rightarrow \frac{t_1}{\sqrt{2}}=t_2$