Electrostatics - Metal Shell Question

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SUMMARY

The discussion centers on a problem involving a thick metal shell with inner radius 2 cm and outer radius 4 cm, containing a point charge of 3 nC at its center. The induced charge on the inner boundary of the shell is -3 nC, while the induced charge on the outer boundary is +3 nC. This conclusion is derived from the principles of electrostatics, specifically the behavior of conductors in electrostatic equilibrium. The participant initially miscalculated the electric field but later recognized the importance of understanding charge distribution within conductors.

PREREQUISITES
  • Understanding of electrostatics and Gauss's Law
  • Familiarity with the properties of conductors in electrostatic equilibrium
  • Knowledge of electric field calculations, specifically using the formula E = q/4∏ ε0 r^2
  • Basic concepts of charge induction and distribution in conductive materials
NEXT STEPS
  • Study Gauss's Law and its applications in electrostatics
  • Learn about electric field calculations in spherical coordinates
  • Explore the concept of charge induction in conductors
  • Review problems involving multiple charges and their effects on electric fields
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electrostatics, as well as educators and anyone preparing for exams related to electric fields and charge distributions in conductors.

Cazicami
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Homework Statement


Consider a thick metal shell with inner radius a = 2cm and b = 4cm. The shell carries no net charge. A point charge of q = 3nC is placed at the centre of the shell. Find the total charged induced on the inner boundary of the shell. Find the total charge induced on the outer boundary of the shell.

5 marks.


Homework Equations


E = q/4∏ ε0 r^2 - I think.


The Attempt at a Solution


My attempt was to use the equation above to work out the charge at each radius. However the answer I get is very wrong. The solutions simply state,
"Obviously, the charge induced on the inner boundary is -3nC, while the charge induced on the outer boundary is 3nC." So this is making me think I have missed something. Considering the value I calculated was E = 67438NC^-1


Any help you can give would be great, as I am studying for my resits.
Thanks
 
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you have calculated the electric field at the inner radius due to just the charge in the middle. But the question does not ask for this. And this is not really important, either. Think about what happens inside a conductor in electrostatic problems. You have probably used this concept before.
 
Thanks, Sorry it was such a noob question, but I have now found it in my notes.

Thanks so much for your help.

Cazi
 

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