Calculating charge on capacitor.

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    Capacitor Charge
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SUMMARY

The discussion focuses on calculating the charge on a capacitor, specifically addressing the uniformity of surface charge density (σs) on a conducting plate. It establishes that for a circular capacitor, the charge (Q) can be calculated using the formula Q = σs[(π)(a²)], where 'a' is the radius. However, it is noted that in practical scenarios, such as with a finite conducting plate, the surface charge density is not uniform due to charge repulsion, leading to variations in charge density, particularly towards the edges. This highlights the complexity of charge distribution in capacitors with non-uniform electric fields.

PREREQUISITES
  • Understanding of electrostatics and charge distribution
  • Familiarity with the concept of surface charge density (σs)
  • Knowledge of calculus, specifically integration techniques
  • Basic principles of capacitor design and function
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  • Study the effects of dielectric materials on charge distribution in capacitors
  • Learn about the implications of non-uniform electric fields in capacitor design
  • Explore advanced integration techniques for calculating charge in complex geometries
  • Investigate the behavior of charge density in finite conducting plates
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Students and professionals in electrical engineering, physicists, and anyone involved in capacitor design and analysis will benefit from this discussion.

Miike012
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The problem and solution are in the document.

My question is in regards to the solution about calculating the charge on the capacitor. Would an equally correct solution for the charge be...
Let σs be the surface charge density on the plate.
The plate is a conductor and I would assume that the thickness throughout the plate is uniform, and there should be no divots or anything that would create the charge density in some area of the plate to be greater than another area of the plate.

Thus σs is uniform.

Therefore Q = ∫sσsds = σs[∫sds] =
σs[(pi)(a2)] where a is the radius of the circular capacitor.
(ds is obviously a differential surface area of the capacitor)

FINAL: Charge of capacitor (Q) = σs[(pi)(a2)]
 

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Yes - if you know the surface charge density.

In the attachment problem, you don't know the surface charge density.

For a finite conducting plate, the surface charge density will not, generally, be uniform: like charges repel so there may be a higher charge density towards the outer edges of the plate. In the attachment problem, the dielectric has a radially dependent constant - so the charge distribution is certainly not uniform (as E is not uniform).
 

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