Question

A point charge +Q having mass m is fixed on horizontal smooth surface. Another point charge having magnitude +2Q & mass 2m is projected horizontally towards the charge +Q from the distance with velocity $V_0$.
Force applied by floor on the fixed charge in horizontal direction, when distance between charges becomes 'd'

• A. $\frac{2K{Q}^2}{d^2}$
• B. $\frac{K{Q}^2}{d^2}$
• C. zero
• D. none

Answer 1

The correct option is A $\frac{2K{Q}^2}{d^2}$

A charge $=+Q$

mass $=m$

magnitude $=+2Q$

mass $=2m$

Velocity $=V_0$

Using coulomb's law, charge $Q$ and thus the force

$F=K\frac{Q^2}{d^2}=0$ and it will become $O$ because it is inner surface.

Now, force applied by floor on the fixed charge in horizontal direction

$F=\frac{K\left(2Q\right)\left(Q\right)}{d^2}=\frac{K\times {2Q}^2}{d^2}=\frac{2K{Q}^2}{d^2}$