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

• Tuatara
Tuatara
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

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

#### Attachments

• Image1.png
2.1 KB · Views: 4
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?

• Introductory Physics Homework Help
Replies
1
Views
809
• Introductory Physics Homework Help
Replies
11
Views
183
• Introductory Physics Homework Help
Replies
17
Views
485
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
914
• Introductory Physics Homework Help
Replies
18
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
378
• Introductory Physics Homework Help
Replies
26
Views
658
• Introductory Physics Homework Help
Replies
1
Views
139