Two charged objects attract each other with a force of F. What happens to the force between them if one charge is doubled, the other charge is tripled, and the separation distance between their centers is reduced to ¼ its original value?

The force is increased by a factor of 96. (F1 = 96F)

Explanation:

Let the first charge be Q1

Let the second charge be Q2

Let the distance between their centers be r

The electrostatic force, F, between them is:

F = (k*Q1*Q2) / r²

If the first charge is doubled = 2Q1

If the second charge is tripled = 3Q2

The separation between their centers is reduced to ¼ = ¼r

The force between them becomes:

F1 = (k* 2Q1 * 3Q2) / (¼r)²

F1 = (6 * k * Q1 * Q2) / (r²/16)

F1 = 96(k * Q1 * Q2) /r²

F1 = 96F

The force is increased by a factor of 96.

u can delete ur account.

1. go on google and write

2 open this and click on your profile pic and if your profile is not appearing, then u have to log in.

3. when u click on your profile pic, there appears four options on the top.

4. click on the 1st option and scroll down and go at the bottom.

5. click on edit profile.

6. there appears a option 'prefernces' and at that option at the right side , 'edit' is written.

7. click on the edit option.

8. there is a option " i want to delete my account" and click on that.

hope this u

and ur welcm

: )