Electric Potential: Solve for Energy to Move -5uC Charge to Infinity

[bob19841984]

Homework Statement

A point charge of +8C is located at (-6,0)cm. A second point charge of -11C is located
at(-5,-12)cm. If a -5C charge were at the origin, how much energy is required to move it to in nity?

Homework Equations

V=KQ/R
U = -q
V

The Attempt at a Solution

I have the solution. It is given as below... I just don't understand why!
http://a-s.clayton.edu/jdyer/physics/phys1112/phys1112S09/Quizzes/1112_Quiz2_solution.pdf"

If the origin now has a -5 micro C charge, why is the energy potential still calculated as if it was not there? I have been banging my head on this for a while, any help would be greatly appreciated.

 

[rl.bhat]

1. T

The Attempt at a Solution

I have the solution. It is given as below... I just don't understand why!
http://a-s.clayton.edu/jdyer/physics/phys1112/phys1112S09/Quizzes/1112_Quiz2_solution.pdf"

If the origin now has a -5 micro C charge, why is the energy potential still calculated as if it was not there? I have been banging my head on this for a while, any help would be greatly appreciated.

V₁ and V₂ are the potentials at origin due to two charges. -5 μC charge is taken from origin to infinity and the work done to do this is -q*ΔV.

 

[nlightner]

I would respond by saying that the solution provided is correct. The reason why the energy potential is still calculated as if the -5 micro C charge is not there is because it is assumed that the charge at the origin is infinitesimally small compared to the charges at (-6,0)cm and (-5,-12)cm. This means that the electric field created by the -5 micro C charge at the origin is negligible and can be ignored. Therefore, the potential energy for the -5 micro C charge at the origin is solely influenced by the electric field created by the +8 micro C charge at (-6,0)cm and the -11 micro C charge at (-5,-12)cm. I hope this explanation helps to clarify any confusion.