# The work done on charges

1. Oct 21, 2012

### Fabio010

Two charges
q1 = 3.0µC and q2 = −4.0µC initially are separated by a distance ro = 2cm.

An external agent moves the charges until they are rf = 5cm apart.

a) How much work is done by the electric field moving the charged from ro to rf? Is it negative or positive? What is the work done by a external agent? Is it positive or negative?

Attempt:

-W = ΔU

Initial U = ke*(q1*q2)/(ro)

Final U = ke*(q1*q2)/(rf)

-W = ke*(q1*q2)/(rf) -ke*(q1*q2)/(ro)

the work is positive.

Because the Wext = -W, then Wext is negative.

I think that my attempt is wrong because in next questions it answers for the potential energy of the initial state....

2. Oct 21, 2012

### TSny

Your equations look good, but not the conclusion. Check the signs again.

3. Oct 21, 2012

### Fabio010

first of all sorry for posting in the wrong section.

because rf > ro

ke*(q1*q2)/(rf) -ke*(q1*q2)/(ro) is going to be a negative number

so W = -ΔU , W is positive.

something is escaping me...?

4. Oct 21, 2012

### TSny

The charges have opposite signs.

5. Oct 21, 2012

### Fabio010

damn... :)

So the work done by the electric field is negative, and the work done by the external agent is positive.

But one more question. The following questions of the exercise are.

c) what is the potential energy of the initial state where the charges are ro = 2 cm apart?

d) what is the potential energy of the initial state where the charges are ro = 5 cm apart?

e) what is the change in potential energy from the initial state to the final state?

Based in the logic of these questions, my answer for the first 2 questions (work done by the electric field and the external agent) is wrong. Don't you think so?

6. Oct 21, 2012

### TSny

So, you've essentially already answered c, d, and e in answering a and b. Two possibilities: (1) maybe they want you to be clear on the signs of each quantity and how they are all related. (2) Maybe they want you to get the work in part (a) by explicit integration of the Coulomb force.

I don't know. But what you've done looks fine to me.

7. Oct 21, 2012

### Fabio010

Ok thanks for the help!