Parallel Plate Capacitor Diameter

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SUMMARY

The discussion centers on calculating the diameter of disks forming a parallel-plate capacitor, given a separation of 0.50 mm and an electric field strength of 3.50 × 105 N/C after transferring 1.70 × 109 electrons. The initial approach incorrectly used the formula for the electric field of a charged disk instead of the appropriate equations for parallel-plate capacitors. To find the correct diameter, participants suggest converting the charge to Coulombs and applying Gauss's law to determine the area of the plates.

PREREQUISITES
  • Understanding of parallel-plate capacitor principles
  • Knowledge of Gauss's law
  • Ability to convert charge from electrons to Coulombs
  • Familiarity with electric field calculations
NEXT STEPS
  • Review the equations for electric fields in parallel-plate capacitors
  • Learn how to apply Gauss's law for calculating electric fields
  • Practice converting charge from electrons to Coulombs
  • Explore the relationship between area, capacitance, and electric field strength
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone involved in capacitor design or analysis.

tomizzo
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Homework Statement



Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. Transferring 1.70 ×109 electrons from one disk to the other causes the electric field strength to be 3.50 ×105N/C.

What are the diameters of the disks?

Homework Equations



Refer to attachment

The Attempt at a Solution



So I substitute everything into the formula and end up getting R = .5 mm. I then double this to find the diameter and still get the wrong answer. Any ideas?
 

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tomizzo said:

Homework Statement



Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. Transferring 1.70 ×109 electrons from one disk to the other causes the electric field strength to be 3.50 ×105N/C.

What are the diameters of the disks?

Homework Equations



Refer to attachment

The Attempt at a Solution



So I substitute everything into the formula and end up getting R = .5 mm. I then double this to find the diameter and still get the wrong answer. Any ideas?

Not sure what you're intending to do with that equation you attached. It looks like the on-axis field for a charged circular disk at some distance z from the disk along the axis. But here you have a parallel plate capacitor and the field will be (essentially) uniform between the parallel plates.

Look instead to formulas related to parallel plate capacitors.
 
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convert Q to Coulombs, then use Gauss to obtain Area.
 

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