Work, energy, power unit (coulombs)

[ferrariistheking]

Homework Statement

For a positively-charged particle at the origin, and a negatively-charged particle lying to its right on the positive x-axis, the electrostatic force on the negative charge is attractive, and Coulomb's Law gives the magnitude of that force:

F = k |Q1| |Q2| / x^2

k is Coulomb's constant. It has a value of 8.99 x 109 N-m2 / C2.

The positively-charged particle has a charge of Q1 = 0.021 Coulombs.

The negatively-charged particle has a charge of Q1 = -0.051 Coulombs.

How much work does the positive charge do on the negative charge as their separation decreases from 0.035 m to 0.018 m?

Homework Equations

F = k |Q1| |Q2| / x2
work = Force x distance

The Attempt at a Solution

I plugged in k and then the two charges. I then did 0.035m -0.018m and got 2.89e-4. The result was F and I then multiplied F with the distance, 2.89e-4.

Does somebody know what I did wrong here?

 

[mfb]

I then did 0.035m -0.018m

What does that mean?

and got 2.89e-4

What is that, and what is its unit?

You can calculate the force both for the larger initial separation and the smaller final separation, but if you move the charge the force will increase in the process. You'll need an integral, or an approach via a useful conservation law.

 

[Abhishek kumar]

Apply conservation of energy for two charge system.change in potential energy is equal to work done.