Question

Two particles $A$ and $B$ having same mass have charge $+q$ and $+4 q$, 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, $B=+4 q$
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
$\left.K=V \times Q \cdots \cdots(ii\right)$
From Eqs. (i) and (ii), we have
$\frac{1}{2} m v^2=V \times Q$
For particle $A$,
$q V=\frac{1}{2} m v_A^2 \cdots \cdots(\text { iii })$
For particle $B$
$4 q V=\frac{1}{2} m v_B^2 \cdots \cdots(i v)$
Dividing Eq. (iii) by Eq. (iv), we get
$ \frac{1}{4}=\frac{v_A^2}{v_B^2}$
$\Rightarrow \quad \frac{v_A}{v_B}=\frac{1}{2}$
Hence, the ratio of their speed $\frac{v_A}{v_B}=\frac{1}{2}$