Question

Two point charges attract each other with an electric force of magnitude F.a. 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?b. If one charge is reduced to one-fourth its original value, how should the distance between the two charges change such that the force F remains constant?

Answer 1

a.
`F'=(1)/(12)F`

b.
`d'=(d)/(2)`

Step-by-step explanation:

a. The magnitude of the electric force is given by the Coulomb's law:

`F=(kq_1q_2)/(d^2)`

In this case, we have
`q_1'=(q_1)/(3)` and
`d'=2d`:

`F'=(kq_1'q_2)/(d'^2)\\F'=(k((q_1)/(3))q_2)/((2d)^2)\\F'=(kq_1q_2)/(3*4d^2)\\F'=(1)/(12)(kq_1q_2)/(d^2)\\F'=(1)/(12)F`

b. In this case, we have
`q_1'=(q_1)/(4)`:

`F=(kq_1'q_2)/(d'^2)\\(kq_1q_2)/(d^2)=(k((q_1)/(4))q_2)/(d'^2)\\(1)/(d^2)=(1)/(4d'^2)\\d'^2=(d^2)/(4)\\d'=(d)/(2)`