Electrostatic force between 2 hemispheres

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SUMMARY

The discussion centers on calculating the electrostatic force of repulsion between the northern and southern hemispheres of a charged metal sphere with radius R and total charge Q. The relevant equation involves integrating the surface charge density σ and the average electric field Eav, specifically using the formula Fz = ∫ fz dA. Participants express confusion regarding the integration process, particularly the distinction between integrating over the surface area versus the volume of the sphere, and the implications of azimuthal symmetry on the force calculations.

PREREQUISITES
  • Understanding of electrostatics, specifically Coulomb's Law
  • Familiarity with surface charge density (σ) and electric fields (E)
  • Knowledge of spherical coordinates and integration techniques
  • Basic principles of mechanics and electromagnetism (EM)
NEXT STEPS
  • Study the derivation of electric fields from surface charge distributions
  • Learn about the integration of forces in spherical coordinates
  • Explore the concept of azimuthal symmetry in electrostatics
  • Review the relationship between mechanics and electromagnetism in physics curricula
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators and anyone looking to deepen their understanding of electrostatic forces and integration techniques in spherical coordinates.

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Homework Statement


A metal sphere of radius R carries a total charge Q, what is the force of repulsion between the northern and southern hemispheres

Homework Equations


\large f=\sigma Eav=1/2 \sigma(Eabove+Ebelow). \\Fz=\int fz dA =\int \sigma Eav cos\theta R^{2}sin\theta d\theta d\phi=\int \frac{Q}{2*4\pi R^2}\frac{Q}{4\pi\epsilon o R^2}cos\theta R^{2}sin \theta d\theta d\phi

The Attempt at a Solution


I'm not sure why you don't integrate the R^2 in spherical coordinates? when you integrate are you summing force differential elements from each piece from the volume of the sphere? also since E is 0 inside the conductor and there is a force only outside, the force tends to separate the hemispheres and then doesn't the force cancel (this seems to be the calculation for one hemisphere only)? maybe I don't understand the integral sum of the force elements?
 
Last edited:
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You're not integrating over the volume; you're integrating over the area. A sphere is a two-dimensional surface.
 
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thanks, sorry so I am summing the differential force elements in the z direction from the surface of the sphere by taking Ez? but aren't there radial force elements in all directions which cancel in the +z -z with azimuthal symmetry? or if you calculate the force of each hemisphere then wouldn't the integration be from 0 to pi for phi instead of 2 pi?

do most people take mechanics before EM like from some text?
 
Last edited:

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