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.

ill do it for 100, also this isnt 80 its 40

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2