Question

Two charged particles P and Q having the same charge $1\mu C$ and mass $4\;kg$ are initially kept at the distance $1\;mm$. Charge P is fixed, then the velocity of charge Q when the separation between them becomes $9\;mm$

& A.3\times {{10}^{3}}m/s \\\ & B.2\times {{10}^{3}}m/s \\\ & C.5\times {{10}^{3}}m/s \\\ & D.7\times {{10}^{3}}m/s \\\ \end{aligned}$$

Answer 1

Electric field is the electric force due to a unit positive charge which is at rest would exert on its surrounding. We also know that the electric potential due to a charge, is defined as the amount of energy needed to move a unit positive charge to infinity. Using the relation between the two we can solve this sum.
Formula used:
$\Delta P.E=KE$

Complete answer:
We also know that the electric potential due to a charge $q$ is defined as the amount of energy needed to move a unit positive charge to infinity. Also the potential at any point is the vector sum of potentials at that point. Also, potential is proportional to the charge and inversely proportional to the square of the distance between the point and the charge. $V=\dfrac{kq}{r}$, where $r$ is the distance between the unit charges and $k=\dfrac{1}{4\pi\epsilon_{0}}$ which is a constant
We know from the energy is conserved, then we can say that the change in potential energy leads to the kinetic energy of the charges.$\Delta P.E=KE$
Given that P and Q have charge $q=1\mu C$ and mass $m=4\;kg$. Also given that $r_{1}=1mm$ and $r_{2}=9mm$, let assume that the charge moves with a speed of $v$. Then we have,
$kq\left(\dfrac{1}{r_{1}}-\dfrac{1}{r_{2}}\right)=\dfrac{1}{2}mv^{2}$
$\implies 9\times 10^{9}\times 10^{-6}\left(\dfrac{1}{10^{-3}}-\dfrac{1}{9 \times 10^{-3}}\right)=\dfrac{1}{2}4v^{2}$
$\implies v^{2}=\dfrac{9\times 10^{9}\times 10^{-6}\left(\dfrac{8}{9 \times 10^{-3}}\right)}{2}$
$\implies v^{2}=\dfrac{8\times 10^{6}}{2}$
$\implies v^{2}=4\times 10^{6}$
$\implies v=2\times 10^{3}m/s$

So, the correct answer is “Option B”.

Note:
The potential difference is also the potential energy possessed by the charges; hence we are using the formula for potential difference to calculate the difference in potential energy. This is the energy of charge due to some position. Also, note that, $E=\dfrac{V}{r}$ where $r$ is the distance of the charge from the unit charge.