Electric field in a spherical shell

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SUMMARY

The discussion centers on calculating the electric field and surface charge distribution for a conducting spherical shell with a -5 nC point charge at its center and a total charge of +7 nC. The electric field at r = 1 m is calculated as -45 N/C, indicating an inward direction. At r = 3 m, the electric field is 0 N/C due to the properties of conductors in electrostatic equilibrium. For r = 5 m, the electric field is determined to be 0.719 N/C, influenced by the total charge on the shell. The surface charge distribution on the outer shell can be derived using Gauss' Law, specifically the equation E = σ/ε₀.

PREREQUISITES
  • Understanding of Gauss' Law
  • Familiarity with electric fields and charge distributions
  • Knowledge of electrostatics in conductors
  • Basic calculus for field calculations
NEXT STEPS
  • Study the application of Gauss' Law in different geometries
  • Learn about electric field calculations in spherical coordinates
  • Explore the concept of electrostatic shielding in conductors
  • Investigate the relationship between surface charge density and electric field
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Students and professionals in physics, electrical engineering, and anyone studying electrostatics, particularly those focusing on electric fields and charge distributions in conductive materials.

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Homework Statement


A -5-nC point charge is located at the center of a conducting spherical shell. The shell has an inner radius of 2 m, an outer radius of 4 m, and a charge of +7 nC. (Let the radially outward direction be positive.)
(a) What is the electric field at r = 1 m? (Indicate the direction with the sign of your answer.)




(b) What is the electric field at r = 3 m? (Indicate the direction with the sign of your answer.)


(c) What is the electric field at r = 5 m? (Indicate the direction with the sign of your answer.)



(d) What is the surface charge distribution, σ, on the outside surface of the shell? (Indicate the direction with the sign of your answer.)




The Attempt at a Solution



a) you get this by doing -5 * 10^-9 * 9 * 10^9 / (1) ^2 = -45 N/C


(b) 0 N/C

i' not sure of this i thin it's because it's on the middle. Middle meaning middle of 2 M and 4 M which is 3M
i would like a clear answer

(c) What is the electric field at r = 5 m? (Indicate the direction with the sign of your answer.)
Units are required for this answer. 0.719 N/C

this i have no idea.. i think there is a formula but not sure how the charge of 7nC affects the field

(d) What is the surface charge distribution, σ, on the outside surface of the shell? (Indicate the direction with the sign of your answer.)

i think you use E = σ/epsilon0
 
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