Question

Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them?a) 3F/2b) F/3c) F/6d) F/12e) 3F/4

Answer 1

New force,
`F'=(F)/(12)`

Step-by-step explanation:

Given that, two point charges attract each other with an electric force of magnitude F. It is given by :

`F=k(q_1q_2)/(r^2)`

If one charge is reduced to one-third its original value and the distance between the charges is doubled such that,

`q_1'=(q_1)/(3)`,
`r'=2r`

`F'=k(q_1'q_2')/(r'^2)`

`F'=k((q_1/3)q_2)/((2r)^2)`

`F'=(F)/(12)`

So, the electric force between them is reduced to (1/12). Hence, the correct option is (d).