How much energy is needed to place 4 positive charges

[shashaeee]

Can someone just confirm if I'm doing these two problems correctly ...

1. How much energy is needed to place 4 positive charges, each of magnitude +5mC, at the vertices of a square of side 2.5cm

2. Choose one way of assembling the charges and calculate the potential at each empty vertex as this set of charges is assembled.


For #1, I used the formula PE = kQQ/r

PE = 9.0 x 10⁶J

and because there are 6 sets (4 sides and 2 diagonals) with similar lengths and charges, I added the PE values together, giving me a total of PE_T = 5.4 X 10⁷J

For #2, I'm not sure what exactly its asking me to do, but I used the formula V = kQ/r

For the r, I used Pythagoreom Theorom to find the distance to the middle

V = 2.55 x 10⁹V

and I'm not really sure, but there are 4 sets this time?

So Total Potential = V_A1+V_A2+V_A3+V_A4?

 

[TSny]

Hi, shashaeee.

For problem 1, did you treat the "diagonal" pairs as having the same distance r as the "side" pairs?

For problem 2, it appears to me that you want to think of assembling the square one charge at a time and at each step calculate the potential at the remaining empty vertices. So, suppose you started by placing one charge at the upper left corner of the square. You would then have 3 empty vertices. So, you should calculate the potential at each of those three empty vertices. Then suppose you add the second charge at the upper right corner. You would be left with 2 empty vertices. So, calculate the potential at those 2 empty vertices, and so on.

 

[shashaeee]

Oh, no I didn't! Thank you!


So for #2 ... basically ...

If my box was arranged as

1 _____3
|
|
2 _____ 4

V₁ = V₂+V₃+V₄

V₁₂ = V₃+V₄

V₁₂₃ = V₄

 

[TSny]

My interpretation of #2 is just to calculate the potential at each empty corner, but don't add them together. So, when charge 1 is in place, you'll calculate V₂, V₃, and V₄. When charges 1 and 2 are in place, you'll calculate new values for V₃ and V₄. When charges 1, 2 and 3 are in place, then you'll recalculate V₄.

At least that's my reading of the problem. Hope I'm not misinterpreting it.