Confusion with Electric Potential/Potential Energy

  • Thread starter sisigsarap
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  • #1
sisigsarap
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There is a problem I am finding quite confusing:

Two particles, with charges of 20.0 nC and -20.0nC, are placed at the points with coordinates (0,4.00cm) and (0,-4.00cm). A particle with charge 10.0 nC is lcoated at the origin. Find the electric potential energy of the configuration of the three fixed charges.

I am having difficulty understanding the last sentence. For electric potential energy I thought I would use V = 8.99*10^9(q/r) , where I would sum up each charge? For example:

V1 = 8.99*10^9(20nC/4cm) V2 = 8.99*10^9(-20nC/4cm) V3=8.99*10^9(10/0)

Then the Electric potential energy would be V1 + V2 + V3 = 0?

Or would I use U = 8.99*10^9(q1*q2)/(distance from q1 to q2)

and follow the same procedure?

I am very confused with the question and what the book refers to as potential energy and electrical potential??

Any help is appreciated!

Thanks!
 

Answers and Replies

  • #2
Andrew Mason
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sisigsarap said:
Two particles, with charges of 20.0 nC and -20.0nC, are placed at the points with coordinates (0,4.00cm) and (0,-4.00cm). A particle with charge 10.0 nC is lcoated at the origin. Find the electric potential energy of the configuration of the three fixed charges.
Since the electric potential, V, is the potential energy per unit charge it is a scalar quantity that is a function of the separation and the charge. Since it is directionless, potentials can just be added together.

[tex]V = U/q = \int_\infty ^R \frac{kQ}{r^2}dr = \frac{kQ}{R} - 0[/tex]

[tex]U = Vq = \frac{kQq}{R}[/tex]

In order to find the total potential energy, take the potential energy of each pair of charges and add them up (be careful with the signs).

The answer, I think, is -4.5e-5 J.

AM
 

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