Question

Two identical small spheres charged to 20nC and –15nC are placed 0.1m apart in air.

Find the force between them. Then they are brought into contact and again separated

by the same distance. Find the new force between them. plz tell fast ​

Answer 1

Explanation:

F=kq1q2/r^2

F=[(9*10^9)*(20*10^-9)*(15*10^-9)]/0.1

F=(2700*10^-9)/0.1

F=27*10-6 N

If bought together--

Charge on both spheres=20-15/2=2.5*10^-9C

New force

F=(9*10^9)*(2.5*10^-9)*(2.5*10^-9)/0.1

F=56.25*10^-9 N