Capacitance of two unequal spheres

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SUMMARY

The capacitance of two unequal charged spheres can be estimated using COMSOL, an electrostatic modeling software. By creating a model that calculates the electric field and potential difference between the spheres, one can derive the capacitance as the charge of one sphere divided by the voltage between them. The process involves drawing a diagram of the electric field, assigning values based on voltage and distance, and performing spherical integrals to determine charge distribution. This method has shown acceptable agreement with experimental results.

PREREQUISITES
  • Understanding of electrostatics and capacitance principles
  • Familiarity with COMSOL Multiphysics software
  • Knowledge of numerical methods for evaluating integrals
  • Basic skills in drawing and interpreting electric field diagrams
NEXT STEPS
  • Explore COMSOL Multiphysics documentation for electrostatic modeling examples
  • Learn about spherical integrals in electrostatics
  • Research numerical methods for electric field calculations
  • Study the impact of charge distribution on capacitance in non-uniform systems
USEFUL FOR

Physics students, electrical engineers, and researchers interested in electrostatics and capacitance calculations involving complex geometries.

peroAlex
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Assume that we have two charged spheres with unequal radii and which do not overlap. How cold we express their capacitance?
 
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peroAlex said:
Assume that we have two charged spheres with unequal radii and which do not overlap. How cold we express their capacitance?
That's a pretty complex problem. What is the application? Is this a schoolwork problem? Do you have access to COMSOL or another modeling software package?
 
berkeman said:
That's a pretty complex problem. What is the application? Is this a schoolwork problem? Do you have access to COMSOL or another modeling software package?

Literary it's my curiosity at work. I know how to compute capacitance for system of two charged spheres with equal radius, but I never came across example of how two charged spheres with unequal radius.

How could we, for example, find such capacitance? I do have COMSOL.
 
To estimate it roughly, you can just draw a picture like this (each end of each line is orthogonal to the sphere):
MokWZ.png
,
then assign to each line a value of electric field, roughly equal to the voltage divided by the line's length;
then draw a rough diagram of the field on the big sphere, depending on the spherical angle,
multiplied by the electric constant, it's the charge distribution;
then, using some math, estimate the spherical integral and get the whole charge.
I did such estimations for similar problems, there was acceptable agreement with experiment.
 
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peroAlex said:
How could we, for example, find such capacitance? I do have COMSOL.

In COMSOL it is almost trivial, it is just an electrostatic model. I believe there is a somewhat similar example in the manual.
 
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If you have the charge stored in the spheres you can find and vectorial equation for the electric field, right?
With this equation you can use numerical methods to evaluate the integral of electrical field through the line which is the smallest path between the spheres (a line) and evaluate the potential field difference (voltage).
The capacitance will be the charge of one of the spheres divided by the voltage between they.
Can someone point any mistake in my thoughts?
 
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AlexCaledin said:
To estimate it roughly, you can just draw a picture like this (each end of each line is orthogonal to the sphere):
MokWZ.png
,
then assign to each line a value of electric field, roughly equal to the voltage divided by the line's length;
then draw a rough diagram of the field on the big sphere, depending on the spherical angle,
multiplied by the electric constant, it's the charge distribution;
then, using some math, estimate the spherical integral and get the whole charge.
I did such estimations for similar problems, there was acceptable agreement with experiment.

This was very helpful. In my free time I did some extra experimentation. Thank you so much for your time and patience!
 

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