Question

In a free space, a particle A of charge $1\mu C$ is held fixed at point P. Another particle B of the same charge and mass $4\mu kg$ is kept at a distance of 1mm from P. If B is released, then its velocity at a distance 9mm from P is: [Take$\dfrac{1}{4\pi {{\in }_{\circ }}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$ ]
$\begin{align}
  & a)2.0\times {{10}^{3}}m/s \\
 & b)3.0\times {{10}^{4}}m/s \\
 & c)1.5\times {{10}^{2}}m/s \\
 & d)1.0m/s \\
\end{align}$

Answer 1

Hint: In the above question we are asked to determine the velocity of particle B at a distance of 9mm from particle A. When a particle moves in space it possesses kinetic energy. When the charge moves between two points in a given region of potential it either gains or loses potential energy. In the above case the charge is moving away from particle A, hence it will lose its potential energy which will be seen in the form of kinetic energy. Therefore from this data we will accordingly determine the velocity of particle B.

Formula used:
$V(r)=\dfrac{Q}{4\pi {{\in }_{\circ }}r}$
$K.E=\dfrac{1}{2}m{{v}^{2}}$
$U=\Delta Vq$

Complete step-by-step answer:
Let us say there exists a charge Q at some point in space. The potential due to that charge at distance ‘r’ is given by,
$V(r)=\dfrac{Q}{4\pi {{\in }_{\circ }}r}$ where ${{\in }_{\circ }}$ is the permittivity of free space. In the above question it is given that particle B is released from 1mm and reaches 9mm from particle A when it is released. This is basically due to potential energy that it had gained initially. If the particle of charge ‘q’ moves across a region of change in potential $\Delta V$ such that it moves from higher to lower potential, the energy(U) lost by it is given by,
$U=\Delta Vq$
This energy is seen in the form of kinetic energy of the particle. If the mass of the particle is ‘m’ and any instant its velocity is ‘v’ then its kinetic energy (K.E) at that instance is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
The particle B moves from a point 1mm to 9mm with respect to particle A. Therefore the loss in potential energy is equal to gain in kinetic energy of particle B. Mathematically this can be represented as,
$\begin{align}
  & \Delta Vq=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow \left( V(1mm)-V(9mm) \right)q=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow \left( \dfrac{q}{4\pi {{\in }_{\circ }}1\times {{10}^{-3}}}-\dfrac{q}{4\pi {{\in }_{\circ }}9\times {{10}^{-3}}} \right)q=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow {{10}^{3}}\dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}}\left( 1-\dfrac{1}{9} \right)=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow \dfrac{8}{9}9\times {{10}^{12}}\times {{(1\times {{10}^{-6}})}^{2}}=\dfrac{1}{2}4\times {{10}^{-6}}kg{{v}^{2}} \\
 & \Rightarrow 8=2\times {{10}^{-6}}{{v}^{2}} \\
 & \Rightarrow {{v}^{2}}=4\times {{10}^{6}} \\
 & \Rightarrow v=2\times {{10}^{3}}m/s \\
\end{align}$
Therefore the correct answer of the above question is option a.

So, the correct answer is “Option A”.

Note: It is to be noted that all the physical quantities have to be expressed in terms of SI units. The basic idea of solving the above question was conservation of energy. Though the above scenario might not be according to reality, if we know how the energy gets transformed into other forms we will be in position to determine the answer more precisely.