Electric Fields and point charges

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SUMMARY

The discussion centers on calculating electric fields and surface charges for a system involving a point charge of -3 pC and two concentric spherical conducting shells with net charges of +22 pC and +18 pC. The participants utilize Gauss's Law and the formula for electric field (E = kq/r²) to derive the charge on the inner and outer surfaces of the shells. Key conclusions include that the inner surface of the smaller shell accumulates a charge of +7 pC, while the outer surface of the smaller shell and the inner surface of the larger shell are determined to be +22 pC and +18 pC, respectively. The electric field calculations at various points are also discussed, emphasizing the importance of understanding electric field behavior in conductive materials.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric field equations (E = kq/r²)
  • Knowledge of charge distribution in conductors
  • Basic concepts of electric flux and surface charge density
NEXT STEPS
  • Study the application of Gauss's Law in electrostatics
  • Learn about electric field calculations for point charges and spherical conductors
  • Explore the concept of electric flux and its relation to surface charge density
  • Investigate the behavior of electric fields within and outside conductive materials
USEFUL FOR

Students and educators in physics, particularly those studying electrostatics, electrical engineers, and anyone involved in understanding electric fields and charge distributions in conductive systems.

  • #31
ya, E=3pC/ (8.854x10^-12)(4pi (.012^2))= 4.68x10^13 N/pC
 
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  • #32
So How do i use this value to now find the charge on the inner surface?
 
  • #33
beachy6 said:
ya, E=3pC/ (8.854x10^-12)(4pi (.012^2))= 4.68x10^13 N/pC

It is wrong. How did you get that number?

ehild
 
  • #34
O i got it, so it would be +3pC
 

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