Calculating Charges and Separation in an Electrostatic Force Problem

[XxXDanTheManXxX]

Hey I am studying for a test, These are one the questions. Anyone know how to solve this?

Two point charges repel each other with a force of 2.0 x 10^-9 N. One of the point charges carries twice the amount of charge carried by the other. When the two charges are moved 0.1 m farther apart from their initial position, the force reduces to 5.0 x 10^-9 N. What are the charges and what was the initial separation between them?

 

[stunner5000pt]

ok since one charge is twice the other then u know the two chares are q1 and 2q1


to start with the force with some separation of D would be
$$F_{1} = 2.0 * 10^{-9}= k \frac{(q_{1})(2q_{1})}{d^2} = k \frac{2q_{1}^2}{d^2}$$

now for part 2
$$F_{2} = 5.0 *10^{-9} = k \frac{2q_{1}^2}{(d+0.1)^2}$$
DONT SOLVE IT YET
One thing u said doesn't make sense if the charges are separated the force will REDUCE
5x10^-9 is BIGGER than 2x10^-9 so perhaps something is wrong in what you typed. In either case it will change the sign for the d+0.1 term that plus sing may turn negative if they were in fact were brought closer which would make sense

 

[XxXDanTheManXxX]

Yes I noticed that, I did type it correct. So would I take this as a mistake in the teachers account?