Electrostatic Potential Energy of a Conducting Sphere

AI Thread Summary
The discussion focuses on calculating the total electrostatic potential energy of a uniformly charged conducting sphere with radius r_0 and total charge Q. The initial approach using U = QV and V = kQ/r led to an incorrect formula. The correct method involves integrating the work done to move a charge from infinity to the sphere's surface, resulting in W = kQ^2/(2r_0). A question about adapting the solution for a charge density instead of total charge was raised, suggesting that the charge density could be incorporated into the integral while keeping the rest of the approach consistent. The conversation emphasizes the importance of proper integration in deriving the potential energy formula.
George3
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Homework Statement


Determine the total electrostatic potential energy of a conducting sphere of radius r_0 that carries a total charge Q distributed uniformly on its surface. Give your answer in terms of Q, r_0, epsilon_0 and appropriate constants.


Homework Equations





The Attempt at a Solution



I know that U = QV and I know that V = kQ/r. I tried to answer it as U = (1/(4pi*epsilon_0))*Q^2/r_0 but that seems to be incorrect. Can anyone point me in the right direction?
 
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The electrostatic potential energy of the sphere is equal to the work done while it is charged.
If there is q charge on the sphere, the potential is kq/r0 on it surface. The work needed to move a charge dq from infinity to the surface of the sphere is:

<br /> <br /> dW=kq*dq/r_0

To get the whole work, you have to integrate from q=0 to q=Q.

ehild
 
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So if you integrate dW = kq *dq/r you end up getting W = k/r * the integral of q *dq from 0 to Q. Which just ends up being W = kQ^2/2r right?
 
George3 said:
So if you integrate dW = kq *dq/r you end up getting W = k/r * the integral of q *dq from 0 to Q. Which just ends up being W = kQ^2/2r right?

Yes, but with r0, the radius of the sphere.

ehild
 
i have a question similar to this. I was wondering if the question was rephrased to say that the sphere has a charge density, p, instead of a charge q how you would answer it?

Would simply become a Q/(volume of sphere) instead of q in your integral equation with everything else remaining the same?

Ps. sorry if this is not the correct format to ask a question (im new on the forum). If you guys want me to make a new thread please let me know thanks!
 
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