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. 1)How will the test charge move immediately after being released?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is the second option

Step-by-step explanation:

Generally the electric force exerted by the charge Q on the charge (q) is mathematically represented as

`F_Q = (kqQ)/(r^2)`

Generally the electric force exerted by the charge 2Q on the charge (q) is mathematically represented as

`F_(2Q) = (kq2Q)/(2r^2)`

Now the net force exerted on q is

`F_(net) = (kqQ)/(r^2) - (2k q Q)/(4r^2)`

`F_(net) = (4kqQ- 2kqQ)/(4r^2)`

`F_(net) = (kqQ)/(2r^2)`

Looking at the resulting equation we see that
`F_(net) > 0`

This implies that the charge q would move to the right