Charge on a grounded conducting sphere in a uniform electric field, after ungrounding and movement

AI Thread Summary
The discussion centers on the behavior of a charged conducting sphere in a uniform electric field created by a parallel plate capacitor. Initially, the sphere is at the electric potential of the top plate, and its charge remains constant after separation from the plate. The potential difference between the top plate and a point on the sphere is calculated using superposition, leading to a total potential at that point. The potential at the top of the sphere is derived from both the capacitor and the sphere itself, resulting in a specific formula. The inquiry concludes with a question about the potential at a point on the sphere opposite to the top point, indicating a need for further clarification.
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Homework Statement
Consider a horizontal parallel plate capacitor and a conducting sphere S of radius R as in the following drawing. S is initially attached to the top plate. It it then detached from the plate and moved tho the middle of the 2 plates
1. Derive the initial electric charge on S.
2. Derive the electric charge and the potential on S right after separation.
3. Derive the difference of potential between the top plate and T, the point at the top of S in its final position.
The sphere is small with respect to the capacitor and should not significantly affect the capacitor.
Relevant Equations
Capacitance of a conducting sphere: C = 4 ∏ε0 R
Charge Q = C V
Image2.png

Question 1:

The sphere is at the electric potential of the top plate. As the sphere is small with respect to the capacitor, one can consider the bottom plate to be at infinity and therefore we can use the capacitance formula as C = 4 ∏ε0 R. The charge Q is therefore Q = C (V -0) = C V.

Question 2:

Right after separation of the sphere from the top plate, the charge Q should remain the same as when attached. As Q and C remain the same, the potential should also remain the same and equal to V.

Question 3:

We’ll calculate the difference of potential between the top plate and the top part of the sphere T by using the superposition of potentials, first with only the parallel plate capacitor and then the potential of the charged sphere on its surface (and also at T).

Parallel plate capacitor potential at T = (d + R) / 2d * V (the potential is linear)

Sphere potential at T remains equal to the initial one V (?)

The total potential at T should be (d + R) / 2d * V + V.

The difference between the top plate and T should be : V – ( ( d + R) / 2d * V + V ) = – ( d + R) / 2d * V
Please help, I’m not quite sure about the result.
 

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Tuatara said:
Question 1:

The sphere is at the electric potential of the top plate. As the sphere is small with respect to the capacitor, one can consider the bottom plate to be at infinity and therefore we can use the capacitance formula as C = 4 ∏ε0 R. The charge Q is therefore Q = C (V -0) = C V.
If the sphere has capacitance C and charge Q then it would be at potential Q/C with no other charges in the vicinity. But there is charge on the top plate and this will contribute to the potential of the sphere.
Tuatara said:
Question 2:

Right after separation of the sphere from the top plate, the charge Q should remain the same as when attached.
Right
Tuatara said:
As Q and C remain the same,
… and nearby charges remain the same…
Tuatara said:
the potential should also remain the same

Tuatara said:
Question 3:

We’ll calculate the difference of potential between the top plate and the top part of the sphere T by using the superposition of potentials, first with only the parallel plate capacitor and then the potential of the charged sphere on its surface (and also at T).

Parallel plate capacitor potential at T = (d + R) / 2d * V (the potential is linear)

Sphere potential at T remains equal to the initial one V (?)

The total potential at T should be (d + R) / 2d * V + V.
With that reasoning, what would be the potential at the point on the sphere diametrically opposite T?
 
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