Question

The electrostatic force of attraction between two charged objects separated by a distance of 1.0 cm is given by F. If the distance between the objects were increased to 5.0 cm, what would be the electrostatic force of attraction between them?

Answer 1

New force between them will become
`(1)/(36)` times

Step-by-step explanation:

Let the charge on both the object are
`q_1\ and\ q_2` and distance between them is is given 1 cm

So r = 1 cm = 0.01 m

Electric force between them is given F

According to Coulomb's between two charges is given by

`F=(1)/(4\pi \epsilon _0)(q_1q_2)/(r^2)=(Kq_1q_2)/(r^2)`

According to question
`F=(Kq_1q_2)/(0.01^2)=(Kq_1q_2)/(10^(-4))`-----------eqn 1

Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m

So new force
`F_(new)=(Kq_1q_2)/(0.06^2)=(Kq_1q_2)/(36* 10^(-4))`------------------eqn 2

Comparing eqn 1 and eqn 2

`F_(new)=(F)/(36)`