Question

Two particles $A$ and $B$ having same mass have charge $+q$ and $+4q$, respectively. When they are allowed to fall from rest through same electric potential difference the ratio of their speeds $v_A$ to $v_B$ will become

• A. $2:1$
• B. $1:4$
• C. $4:1$
• D. $1:2$

Answer 1

Answer: (D)

Given, mass of both particle $=m$
Charge of particle, $A=+q$
Charge of particle, $B=+4q$
Potential difference $=V$
Kinetic energy is given by
$K=\frac{1}{2}m{v}^2\cdots \cdots (i)$
This energy is equal to electrostatic potential energy is
$K=V\times Q\cdots \cdots (ii)$
From Eqs. (i) and (ii), we have
$\frac{1}{2}m{v}^2=V\times Q$
For particle $A$,
$qV=\frac{1}{2}m{v}_A^2\cdots \cdots (\text{ iii })$
For particle $B$
$4qV=\frac{1}{2}m{v}_B^2\cdots \cdots (iv)$
Dividing Eq. (iii) by Eq. (iv), we get
$\frac{1}{4}=\frac{v_A^2}{v_B^2}$
$\Rightarrow \frac{v_A}{v_B}=\frac{1}{2}$
Hence, the ratio of their speed $\frac{v_A}{v_B}=\frac{1}{2}$