Question

Two equal and opposite charges are placed at a certain distance.The force between them is f.If 25% of one charge is transferred to other,then the force between them is

Answer 1

f=kq^2/r^2

if 25%of one charge is transferred to other,

q2=125q/100

q1=75q/100

f'=kq1q2/r^2

15f/16

Answer 2

, the new electrostatic force will be,

F=(k*5q/4*3Q/4)/r^2=15/16f

Electrostatic static force is the force between two charges seperated by a distance.

For example, lighting is an example of electrostatic force.

The electrostatic force is given by,

f=KqQ/r^2

Where, k = constant

Q,q= charges

r=distance between them

Now, according to the question,

25% of one charge is transferred to other.

Hence, q=125q/100=5q/4

Q=75Q/100=3Q/4

Therefore, the new electrostatic force will be,

F=(k*5q/4*3Q/4)/r^2=15/16f