Question

In free space, a particle $A$ of charge $1\mu \text{C}$ is held fixed at a point $P$. Another particle $B$ of the same charge and mass $4\mu \text{g}$ is kept at a distance of $1\text{mm}$ from $P$. If $B$ is released, then its velcoity at a diatnce of $9\text{mm from}P$ is
$[\text{Take}\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}]$

• A. $2.0\times {10}^3\text{m/s}$
• B. $3.0\times {10}^4\text{m/s}$
• C. $1.5\times {10}^2\text{m/s}$
• D. $1.0\text{m/s}$

Answer 1

The correct option is A $2.0\times {10}^3\text{m/s}$

Given:
Charge on $A$, $q_1=1\mu \text{C}$
Charge on $B$, $q_2=1\mu \text{C}$
Mass of $B$, $m_2=4\mu \text{g}$
Initial distance between $(A$ and $B)=1\text{mm}$
After the release of $B$ distance $=9\text{mm}$

Using conservation of energy

$U_i=U_F+\frac{1}{2}m{v}^2$

$\frac{k{q}_1{q}_2}{r_1}=\frac{k{q}_1{q}_2}{r_2}+\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}m{v}^2=k{q}_1{q}_2[\frac{1}{r_1}-\frac{1}{r_2}]$

$v^2=\frac{2k{q}_1{q}_2}{m}[\frac{1}{r_1}-\frac{1}{r_2}]=\frac{2\times 9\times {10}^9\times {10}^{-12}}{4\times {10}^{-6}\times {10}^{-3}}[1-\frac{1}{9}]=4\times {10}^6$

$v=2\times {10}^3\text{m/s}$

Hence option $\left(C\right)$ is correct.