Electrostatic potential energy of a cubical configuration

In summary, the Electrostatic potential energy of a cubical configuration of point charges as shown in the figure is 1.65E-24 J.
  • #1

Homework Statement

Find the Electrostatic potential energy of a cubical configuration of point charges as shown in the figure. Each of the charges is 5.00e and the edge of the cube is 2 cm. (The image is simply a cube with one of the points labeled q)

Homework Equations


The Attempt at a Solution

I tried to do this problem by finding the PE at 3 different distances, being the closest 3 points, which are 2 cm away, 2nd closest 3 points which are sqrt(8) cm away and the farthest point which is sqrt(12) cm away. Then I used the formula to find the PE from each charge and added them all up. I keep getting the answer 1.65E-24 J, yet it's not right. Any help would be much appreciated.
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  • #2
Hello Lamp, welcome to PF :)

You could follow a (very!) similar thread here, but I wonder if Oro's hint really helps you.

How would you go about with the same question for e.g. the 4 point charges on the bottom plane ?
  • #3
You found the PE to put that one (last) corner charge in place (from infinity).
How much PE did it take to put the 1st charge in its corner?
but the second charge DID take Energy to put in place.
Symmetry says, multiply the last charge's PE by the number of charges, then divide by 2.
  • #4
Sorry, I'm still super confused...I don't see how symmetry plays into this and how to take into account the charge of the particle I'm calculating from, sorry!
  • #5
you used PE = U = kQq/r a bunch of times (3 + 3 + 1, right?),
and added all these Energies up. You used Q = 5e, and also used q=5e (right?) => PE = 1.65E-24J
... which is the PE for the last corner charge (q), being influenced by the other 7 charges Q that were already assembled.

Have you seen a Capacitor being charged? the first charge goes in very easily, because there's no E-field, so it takes no Work (0).
The last charge to go in takes eVf of Work, because the potential is going to is Vf
... but the capacitor only stores ½ Qf Vf = Qf Vavg ... just like a spring PE function.
  • #6
So in this case, I would take the PE i got for the last charge (1.65E-24 J), multiply it by 8 and divide by 2? Why does the capacitor only store that much?
  • #7
why does a spring only store ½ Fmax xmax ? because the initial stretch took zero Force.
Work accumulates ... really, it is Faverage xtotal , that is stored as Energy = ½ k x2

Why is d = ½ a t2 ? although vfinal = a t , we know that d = vavg t = ½ vfinal t.
You DO remember the formula for KE? ½ m v2 ... it is the same ½ , from the same averaging process.
Every time a variable is squared (in an accumulating process).

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