Electrical Potential Energy

[rbraunberger]

Homework Statement

Two point charges Q1 = 3.3 µC and Q2 = 7.6 µC are initially very far apart. They are then brought together, with a final separation of 2.6 m. How much work does it take to bring them together?

Homework Equations

$$\Delta$$PE = ( k Q1 Q2 ) / r

Work = - $$\Delta$$PE

k= 8.99 e 9

The Attempt at a Solution

PE final = k Q1 Q2 / r
= (8.99e9)(3.3e-6)(7.6e-6) / ( 2.6m)
= 0.0868 J

PE initial = k Q1 Q2 / r
with r initial being infinitely large PE initial is basically 0

so $$\Delta$$PE = 0.0868 + 0
Work = - $$\Delta$$PE = -0.0868What is the error in this? Thank you!

 

[ehild]

Is the separation 2.6 m or 2.8 m? And do not forget the unit of work!ehild

 

[rbraunberger]

It was 2.6m...I fixed it. I found that the answer was just 0.0868 not -0.0868. My teacher said the equation was Work = -$$\Delta$$PE, but I found it some place else that shows it as just Work = $$\Delta$$PE. Which is correct?

 

[ehild]

The work needed to bring the charges together was the question. It is some external force Fe (maybe your force) that does this work. The external force is opposite and at least equal in magnitude with the Coulomb force Fe=-Fc. As the charges repel each other, the external force must push them toward the centre, so the direction of the external force is the same as the displacement: the work is positive.

ehild

 

[rwisz]

There is no error in this solution. You have correctly used the formula for electrical potential energy and calculated the work required to bring the two point charges together. However, it should be noted that this is the work done by the external force in bringing the charges together, and not the total work done, as there may also be work done by the electric field of the charges themselves. Additionally, it is important to include the units in your final answer, so the correct answer would be -0.0868 J.