Spherical Capacitor: Solving Electric Field, Displacement Field & Capacitance

Click For Summary
SUMMARY

The discussion focuses on solving the electric field (E), displacement field (D), and capacitance of a spherical capacitor with inner radius R and outer radius 3R, containing two dielectrics with permittivities e1 and e2. The electric fields in each region are derived using the formula D = (Einitial)(e) and Gauss's Law, resulting in E1 = q/(4e1pir^2) and E2 = q/(4e2pir^2). The capacitance is calculated as Ctotal = 24piRe1e2/(6e2-e1). The final part of the problem, concerning the surface charge on the e1 dielectric at radius R, is approached using Gauss's Law.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric fields and displacement fields
  • Knowledge of capacitance calculations for capacitors
  • Concept of dielectric materials and their permittivities
NEXT STEPS
  • Study the application of Gauss's Law in various geometries
  • Learn about the properties and calculations involving dielectrics in capacitors
  • Explore advanced capacitance formulas for complex capacitor configurations
  • Investigate the relationship between electric field strength and surface charge density
USEFUL FOR

Students and professionals in electrical engineering, physicists dealing with electrostatics, and anyone studying capacitor design and analysis.

discoverer02
Messages
138
Reaction score
1
I'm stuck on the last part of the following problem:

A spherical capacitor has inner radius R and outer radius 3R, and contains to dielectrics, of permittivities e1, which extends from R to 2R, and e2, which extends from 2R to 3R. Assume charges of +q and -q on the inner and outer surfaces.

a) find E (electric field) and D (displacement field) in each region as functions of r.

b) find the capacitance of the capacitor.

c) what is the surface charge on the e1 dielectric at radius R?

for part a) D = (Einitial)(e) = (kq/r^2)e = q/(4pir^2) for both regions.

and E1 = D/e1 = q/(4e1pir^2)
E2 = D/e2 = q/(4e2pir^2)

for part b) V1 = q/(4piRe1) and V2 = Q(24piRe2)

so C1 = 4piRe1 and C2 = 24piRe2 and Ctotal = 24piRe1e2/(6e2-e1)

Having found all this I don't quite know how to approach part c)

A nudge in the right direction would be greatly appreciated.

Thanks
 
Physics news on Phys.org
Originally posted by discoverer02
A nudge in the right direction would be greatly appreciated.
Apply Gauss's Law.
 
Again, the answer was right in front of my face.

Thanks Doc Al for pointing it out.

Up to this point I've used Gauss's Law only to find E, so since I already had E, I didn't think to use it to find Q.

I need to loosen up, open my mind and not be so intimidated when doing these problems.

Thanks again.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
2K
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
6
Views
1K