Question

If two like charges each of 1c are placed at a distance of 0.3m apart in air then the force between them is

Answer 1

Answer:

Initially 2nC and −8nC are separated by a distance d.

Electrostatic force acting between them initially  F=d2kq1q2

∴  F=d2(2)(8)k=d216k

Now the two charges are allowed to touch each other, they share equal charges among themselves.

Total charge of the system Q=2−8=−6nC

Thus new charge on each sphere q1′=q2′=−3nC

Let the new distance between them be d′.

Thus force acting between them F′=(d′)2kq1′q2′

∴ F′=(d′)2k(−3)(−3)=(d′)29k

But   F′=F

∴ (d′)29k= d216k

Or  (d′)2=169d2

⟹  d′=43d