Electrostatic Potential Energy stored outside a shell.

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SUMMARY

The discussion focuses on calculating the electrostatic potential energy stored outside an isolated conducting shell with a point charge +Q at its center and an additional uniformly distributed charge +Q on its surface. The induced charge on the inner surface is -Q, resulting in a total charge of +2Q on the outer surface, leading to the electric field E=2kQ/r². The formula for energy density, dU/dV=(1/2)ε₀E², is clarified as an energy density rather than the total energy of a charged capacitor, emphasizing the need to apply Gauss's law for accurate calculations.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with Gauss's law
  • Knowledge of energy density concepts in electromagnetism
  • Basic calculus for integration in physics
NEXT STEPS
  • Study Gauss's law applications in electrostatics
  • Learn about energy density in electric fields
  • Explore the derivation of the formula dU/dV=(1/2)ε₀E²
  • Investigate the relationship between electrostatic potential energy and capacitors
USEFUL FOR

Students and educators in physics, particularly those studying electromagnetism, as well as anyone interested in understanding electrostatic potential energy in conducting shells.

cheapstrike
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Homework Statement



A point charge +Q is placed at the centre of an isolated conducting shell of radius R. Find the electrostatic potential energy stored outside the spherical shell if the shell also contains a charge +Q distributed uniformly over it.

Homework Equations


[/B]
E=kQ/r2.

dU/dV=(1/2)εoE2, where V is volume, U is potential energy.

The Attempt at a Solution



The charge +Q inside the conductor will induce -Q charge on the inside surface, which further leads to +2Q charge on the outer surface of the shell. Therefore E=2kQ/r2.

My question is, why have we used the formula dU/dV=(1/2)εoE2. How did we derive it? Isn't it equal to the energy of a charged capacitor.

Sorry if this a bad question, I just want to know how did we derive this formula.

Thanks.
 
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cheapstrike said:
My question is, why have we used the formula dU/dV=(1/2)εoE2
For the same reason that we use mass density ρ to find the mass of an extended object by integration.
cheapstrike said:
How did we derive it?
That you can find in any intermediate-level book on Electricity and Magnetism.
cheapstrike said:
Isn't it equal to the energy of a charged capacitor.
No, it is not. It is an energy density, i.e. energy per unit volume (Joules/m3 in SI units). To do this problem you need to use Gauss's law to find the electric field outside the shell, find the energy density and integrate that over all space outside the shell.
 
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