Question

A thundercloud has an electric charge of 43.2 C near the top of the cloud and -38.7 C near the bottom of the cloud. The magnitude of the electric force between these two charges is 3.95 x 10 ^6 N. What is the average separation between these charges?

HELP??

Answer 1

Do you kno coulombs law?

Answer 2

No, I don't..

Answer 3

Then u should first learn coulombs law

Answer 4

What is it?

Answer 5

go check your textbook :P

Answer 6

Okay, thanks

Answer 7

Yeah, i still don't get it...

Answer 8

coulombs law : F = k*q1q2 / r^2
F = force ,
k = constant = 9*10^9N m2/C2
q1 and q2 = charge on particles in Coulomb's

Note that the charge is MAGNITUDE in strength, and since q1 has a positive charge of 43.2C and q2 has a negative 38.7C this is a attraction force between the two particles, but if it was two positive particles it would repel with the same force in magnitude. The reason I say to note this is in the equation yo have to find the r^2 in distance and if it is negative you can't take the square root of a negative number. So I'm changing the q1 and q2 to both positive values. They will now repel each other, but it will equal the same force if they were attracting each other. Use equation and solve for r:

F = k*q1q2 / r^2
3.95 x 10 ^6 N = [(9*10^9)(43.2C)(38.7C)]/ r^2
r = 1951.73 meters average distance between charges

Answer 9

So it looks like

3.95 x 10^6N = (9*10^9)(43.2C)(38.7C) / 1951.73 ?

Answer 10

the equation I used : F = k*q1q2 / r^2
given , F = 3.95 x 10 ^6 N
substitute in the value for F:
3.95 x 10 ^6 N = [(9*10^9)(43.2C)(38.7C)]/ r^2
Solve the equation for r, which is the only unknown
r^2 = (9*10^9)(43.2)(38.7) / 3.95 x 10^6

The answer is , r = 1951.73 meters, average distance between charges

Answer 11

Thank you @biire2u I get how you did that