Field in between two charged plates in a semi spherical arrangement

In summary, The problem involves two half-spherical electrodes that produce a spherically symmetric electric field. The specifications include the distance between the electrodes, the radii of each electrode, and the voltage across them. The homework equation for this problem is F=q*E (force, charge, electric field). The student is unsure of how to approach the problem, but suggests two possible solutions involving treating the electrodes as point charges or using the relationship between electric field and potential. Another student suggests using a Gaussian shell to compute the charge needed to satisfy the given voltage.
  • #1
Lillensassi
5
0

Homework Statement


We have two half-spherical electrodes, arranged so that they produce a spherically symmetric electric field. What is the magnitude of the electrical force on an electron between the two electrods?

Specifications:
Distance between electrodes: d=0.04 m
Radius for first electrode: r=0.13 m
Radius for second electrode: R=0.17 m
Voltage across the electrodes: V

Homework Equations


F=q*E

(force, charge, electric field)
So what I need to know is the electric field between the plates.


The Attempt at a Solution


So I have basically had two lines of thought. One is to think of it as an entire sphere, in which case there is no electric field due to the outer electrode and just see the inner one as a point charge, to yield the Coulomb force. I am not certain this is okay for a half sphere though, and I cannot find any other deravation for the whole sphere case that Gauss's law and that doesn't work here since it's not a symmetric case (and I only know symmetric cases).

The other though is to somehow relate the electric field to the potential. I mean, I know how to do that for some standard arrangements like point charge or between two sheets. But between half spheres? Is it possible to approximate them with sheets?
 
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  • #2
Your first idea should work because the problem states that the field is spherically symmetric.
 
Last edited:
  • #3
I'd go with a Gaussian shell located between the two hemispherical shells. Compute what Q has to be to satisfy V, then E follows.
 
  • #4
What is a Gaussian shell? Don't the Gaussian surfaces have to be closed?
 
  • #5
Lillensassi said:
What is a Gaussian shell? Don't the Gaussian surfaces have to be closed?

You can close it on the bottom (max. diameter).
 

1. What is the purpose of studying the field in between two charged plates in a semi-spherical arrangement?

The purpose of studying this field is to understand how electric fields behave in various arrangements and how they affect the movement of charged particles. This knowledge is important in fields such as electronics and physics.

2. How is the electric field strength calculated in this arrangement?

The electric field strength can be calculated using the formula E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates. This formula applies to both flat and curved plates in a semi-spherical arrangement.

3. What factors affect the strength of the electric field in this arrangement?

The strength of the electric field is affected by the distance between the plates, the magnitude of the charges on the plates, and the shape of the plates. The field is stronger when the plates are closer together, when the charges are larger, and when the plates are more curved.

4. How does the electric field change as you move from the center of the semi-spherical arrangement towards the edge?

The electric field strength decreases as you move from the center towards the edge of the semi-spherical arrangement. This is because the distance between the plates increases, causing the electric field to spread out and become weaker.

5. How does the presence of a dielectric material between the plates affect the electric field in this arrangement?

The presence of a dielectric material between the plates reduces the strength of the electric field. This is because the material has a lower permittivity than air, which decreases the electric field strength. Additionally, the presence of the material can also lead to a change in the shape of the electric field lines.

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