Electrostatics 2 Question 1
1. In free space, a particle $A$ of charge $1 \mu C$ is held fixed at a point $P$. Another particle $B$ of the same charge and mass $4 \mu g$ is kept at a distance of $1 mm$ from $P$. If $B$ is released, then its velocity at a distance of $9 mm$ from $P$ is
Take, $\frac{1}{4 \pi \varepsilon _0}=9 \times 10^{9} N-m^{2} C^{-2}$
(Main 2019, 10 April II)
(a) $1.5 \times 10^{2} m / s$
(b) $3.0 \times 10^{4} m / s$
(c) $1.0 m / s$
(d) $2.0 \times 10^{3} m / s$
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Answer:
Correct Answer: 1. (b)
Solution:
- Given situation is shown in the figure below,
When charged particle $B$ is released due to mutual repulsion, it moves away from $A$. In this process, potential energy of system of charges reduces and this change of potential energy appears as kinetic energy of $B$.
Now, potential energy of system of charges at separation of $1 mm$ is
$$ \begin{aligned} & U _1=\frac{K q _1 q _2}{r} \\ & q _1=q _2=1 \times 10^{-6} C \\ & \therefore \quad U _1=\frac{9 \times 10^{9} \times 1 \times 10^{-6} \times 1 \times 10^{-6}}{1 \times 10^{-3}}=9 J \end{aligned} $$
Potential energy of given system of charges at separation of $9 mm$ is
$$ U _2=\frac{K q _1 q _2}{r}=\frac{9 \times 10^{9} \times\left(1 \times 10^{-6}\right)^{2}}{9 \times 10^{-3}}=1 J $$
By energy conservation,
Change in potential energy of system of $A$ and $B$
$$ =\text { Kinetic energy of charged particle } B $$
$$ \Rightarrow \quad U _1-U _2=\frac{1}{2} m _B v _B^{2} $$
where, $m _B=$ mass of particle $B=4 \mu g$
$$ =4 \times 10^{-6} \times 10^{-3} kg=4 \times 10^{-9} kg $$
and $v _B=$ velocity of particle $B$ at separation of
$9 mm$
$$ \begin{aligned} & \Rightarrow \quad 9-1=\frac{1}{2} \times 4 \times 10^{-9} \times v _B^{2} \\ & \Rightarrow \quad v _B^{2}=4 \times 10^{9} \Rightarrow v _B=2 \times 10^{3} ms^{-1} \end{aligned} $$
