Q Two Identical coins having similar charges are placed 4 5 m apart on a table Force of - Science - Electricity

Question

Q Two Identical coins having similar charges are placed
4 5 m apart on a table Force of repulsion between them is 49/9 N
The value of charge on each coin is :-
1) 100 mC
2) 200 mC
3) 300 mC
4) 35√10 C
[Explain properly with a solution]

Suraj Raman · Accepted Answer

So, according to Coulomb's Law,
Kq2/r2=49/9
r = 4 5m
K = 9*109Nm2/C2
q2=9×109×q2r2=499=> q2=49×4
529×9×109q = 7×4 5×9×104×10=1
11×10-4

Q. Two Identical coins having similar charges are placed 4.5 m apart on a table. Force of repulsion between them is 49/9 N. The value of charge on each coin is :- 1) 100 mC 2) 200 mC 3) 300 mC 4) 35√10 C [Explain properly with a solution]

Q. Two Identical coins having similar charges are placed 4.5 m apart on a table. Force of repulsion between them is 49/9 N. The value of charge on each coin is :-

1) 100 mC

2) 200 mC

3) 300 mC

4) 35√10 C

[Explain properly with a solution]