# Working with electrostatic force - unknown charge problem

A positive charge of 2X10-12 Coulombs placed 10cm from another charge experiences a repulsive force of 9X10-9 Newtons. What is the sign and magnitude of the other charge?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Since the charge experiences a repulsive force, the sign of the other charge is positive.

(same charges repells)

we have, F = AQ1q2/R^2 .....(1)

where A = 9x10^9

F = 9x10-9 N

Q1= 2x10-12 C

R = 10 cm = 10x10^-2m = 0.1m

R^2 = 0.01 m

from equation1 we get, q2 = F x R^2/A Q1

Substitute the values of the parameters,

q2 = [9x10-9 x0.01]/2x10-12 x 9 x 10^9

= (1/2) x 10^-11 x 10^12 x 10^-9

= 0.5 x 10^-8 Coulomb

The answer to the last posting is as follows sorry for the mistake

The charge of a proton is e= 1.6 x10^-19 coulomb

The force between two charges F = AQ1q2/R^2

where A = 9x10^9

Here we have for this equation,

1*10^-17 = 9x10^9 x 1.6 x10^-19 * 1.6 x10^-19 /R^2

therefore, R^2

= [9x10^9 x 1.6 x10^-19 * 1.6 x10^-19]/1*10^-17

=2.56 x 9 x 10^9 x 10^-21

Or the distance R = sqrt(25.6x10^-30) m

=4.8x10^-15 m

Answer

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