Calculating Electrostatic Pressure on a Charged Sphere

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Homework Help Overview

The problem involves calculating the electrostatic pressure on a charged sphere with a given surface charge density. The original poster attempts to understand how to compute the electric field and the resulting pressure from a small surface area on the sphere.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using Gauss' law to find the electric field and explore the relationship between force, area, and pressure. Questions arise about how to account for the electric field generated by the small surface area when it is removed from the sphere.

Discussion Status

Some participants have provided guidance on calculating the pressure and clarifying the distribution of charge on the small surface area when it is removed from the sphere. Multiple interpretations of the electric field's behavior are being explored, particularly regarding the effects of removing the small surface area.

Contextual Notes

There is a focus on understanding the implications of charge distribution and the assumptions made about the electric field's constancy. The discussion reflects the constraints of the problem as a homework assignment, emphasizing the need for conceptual clarity rather than direct solutions.

springo
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Homework Statement


A sphere with radius R is charged with surface density σ. The charge found in a small surface Δs is repelled by the rest of the sphere, thus generating an electrostatic pressure. Find the pressure.

Homework Equations



The Attempt at a Solution


I think I have to compute the field for the sphere, which can be done using Gauss' law. For Δs: Esph = σ/ε0
Then I think I have to find the field for Δs and subtract it, but I don't know how to find it.
If field is constant, ten just multiplying QΔs = Δs·σ by the field would give the force.

Thanks for your help!
 
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springo said:
1.
If field is constant, ten just multiplying QΔs = Δs·σ by the field would give the force.

Yes. And force per unit area will be pressure.
 
rl.bhat said:
Yes. And force per unit area will be pressure.
Thanks.

Now how do I find the field for Δs?
 
The magnitude of the electric field at every point on the sphere is the same and is equal to Esph = σ/ε0
 
rl.bhat said:
The magnitude of the electric field at every point on the sphere is the same and is equal to Esph = σ/ε0
Yes, but I what meant to is: shouldn't I find the field generated by the sphere and subtract the field generated by Δs to have the field generated by a sphere without Δs? If so, my problem is that I don't know how to find the field generated by Δs.
 
Bump...
 
A body does not exert force on its own surface. So if you want to find pressure on delta s,
assume that it is removed from the sphere and kept very close to the sphere. When you remove delta s from the sphere, charge on each surface of delta s will be 1/2*sigma*delta s. Since delta s is very small, field due sphere close to its surface does no change much. So it will be sigma/ epsilon(not). So the pressure =force /area = 1/2*sigma*delta s*sigma/epsilon(not) = (sigma)^2/2epsilon(not)
 
OK thanks, I understand now. Just one small thing, why is the charge 1/2·Δs·σ? I thought the charge equals to surface (Δs) multiplied by the density (σ). Where does the 1/2 come from?
 
When the delta s is on the sphere , charge sigma delta s is only on outer surface. When you remove it from the sphere, charge distributes equally on both the surfaces. Hence 1/2*sigma*delta s.
 
  • #10
OK thanks!
 

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