Question

What is the repulsive force between two pith balls that are 11.0 cm apart and have equal charges of −38.0 nC?

Answer 1

`F=1.07* 10^(-3)\ N`

Step-by-step explanation:

Given that,

Charge on two balls,
`q_1=q_2=-38\ nC=-38* 10^(-9)\ C`

Distance between charges, r = 11 cm = 0.11 m

We need to find the repulsive force between balls. Mathematically, it is given by :

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

`F=9* 10^9* ((38* 10^(-9))^2)/((0.11)^2)`

F = 0.001074 N

or

`F=1.07* 10^(-3)\ N`

So, the magnitude of repulsive force between the balls is
`1.07* 10^(-3)\ N`.