# Homework Help: Question about 2 charges

1. Aug 6, 2008

### Velocity.

Two charges q1 and q2 are 18cm apart and have an electric force between them of 1micro newton. If the charge on q2 is increased by 9 times and the distance between q1 and q2 changed to 0.621cm, what is the new electric force between them?

2. Aug 6, 2008

### Defennder

You only need the second equation. You aren't given q1q2, so you have to work that out from the first setup of the question. Then simply plug in all the calculated values to get the answer.

3. Aug 6, 2008

### Velocity.

How would I work out q1q2? I'm really stuck and don't know how to proceed

4. Aug 6, 2008

### Defennder

You use the second formula for the force and substitute in the values given.

5. Aug 6, 2008

### Velocity.

Do you mean substitute in 0.18 for r and 1x10^-6 for F and 9x10^9 for k then I get 3.6x10^-18 = q1q2. What do I do next?

6. Aug 6, 2008

### Defennder

Well just calculate the force due to the new setup as described.

7. Aug 9, 2008

### Velocity.

I'm still confused as to how to get the anwer of 7560$$\mu$$N

8. Aug 9, 2008

### Defennder

You only need Coulomb's law for this, nothing else. Look at the formula you're given and think of how to make use of it.

9. Aug 9, 2008

### HallsofIvy

This is really a "proportion" problem. F=kq1q2/r^2 tells you that the force is directly proportional to each charge and inversely proportional to the square of the distance between the charges. You are told that the force, initially, is 1micro newton. If the charge on q2 is multiplied by 9 then so is the force. If the distance between them is changed from 18 cm to 0.621cm, or multiplied by .621/18, then the force is multiplied by (.621/18)2.

I see one ambiguity: I have interpreted "the charge on q2 is increased by 9 times" to mean it is multiplied by 9. You could also interpret "increased by 9 times" to mean 9 times the original charge is add; in other words, the charge is multiplied by 10.