Question

A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q. How will the test charge move immediately after being released?
to the left
to the right
stay still
other
Question) 2) Briefly explain your reasoning

Answer 1

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

            F = $\frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}$

               = $\frac{4KqQ - 2KqQ}{4r^{2}}$

               = $\frac{KqQ}{2r^{2}}$

           F = $\frac{KqQ}{2r^{2}}$ > 0

Thus, we can conclude that the test charge move to the right immediately after being released.