electric field?

Question

i cant understand!,huhu

problem: two charges of equal magnitude exert an attractive force of 4.0 x 10^-4 on each other. If the magnitude of each charge is 2.0 uC, how far apart are the charges?

thanx.

Answer 1

Ok, just remember the Coulombs law :

Fe = K*q1*q2 / d^2

Fe = electric Force

K = constant

q1 and q2 = charges

d = distance

4*10^-4 = 9*10^9*2*10^-6*2*10^-6 / d^2

d^2 = 90

d = sqrt(90) = 9.48 m

Hope that helped

Answer 2

You just need to use Coulomb's Law for the problem (I'm assuming that 2.0 uC is 2.0 micro coulombs):

F = k x (Q1 x Q2) / r^2, where F is the attractive force (4 X 10^-4 N), k is the electrostatic constant (9 x 10^9 N*m^2 / C^2), Q1 and Q2 are the charges of the bodies (2.0 x 10^-6 C) and r is the distance between the bodies (the one you're looking for)

So here's your equation:

r = square root [ (k x Q1 x Q2) / F ]
r = s.r. [ (9x10^9 x 2.0x10^-6 x 2.0x10^-6) / (4x10^-4) ]
r = 9.487 m