Electrostatic Potential Energy stored outside a shell.

In summary: The reason you use a formula like that is that the energy of a charged capacitor is only a special case of this general formula.In summary, to find the electrostatic potential energy stored outside an isolated conducting shell with a point charge placed at its center, we can use the formula dU/dV=(1/2)εoE2, where E is the electric field and εo is the permittivity of free space. This formula is derived from Gauss's law and can be used to calculate the energy density, which can then be integrated over all space outside the shell to find the total energy stored. This formula is not equivalent to the energy of a charged capacitor, but rather a general formula for energy density in electrostatics
  • #1
cheapstrike
42
1

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.
 
Physics news on Phys.org
  • #2
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.
 

1. What is electrostatic potential energy?

Electrostatic potential energy is the energy that is stored in an object due to its position in an electric field. It is the energy that is required to move a charged object from one point to another in an electric field.

2. How is electrostatic potential energy calculated?

The electrostatic potential energy is calculated using the equation U = kQ1Q2/r, where k is the Coulomb constant, Q1 and Q2 are the charges of two objects, and r is the distance between them. The larger the charges and the smaller the distance between them, the greater the electrostatic potential energy.

3. What does it mean to have electrostatic potential energy stored outside a shell?

When an object is surrounded by a charged shell, the electrostatic potential energy can be stored outside of the shell as well as inside. This means that there is a potential difference between the inside and outside of the shell, and the energy is stored in the form of electric fields.

4. How does the electrostatic potential energy outside a shell affect the object inside?

The electrostatic potential energy outside a shell creates an electric field that exerts a force on any charged objects inside the shell. This force can either attract or repel the object, depending on its charge and the charge of the shell.

5. Can electrostatic potential energy outside a shell be changed?

Yes, the electrostatic potential energy outside a shell can be changed by altering the charge or distance of the objects involved. The energy can also be changed by adding or removing charged objects from the electric field or changing the shape of the shell itself.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
354
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
201
  • Introductory Physics Homework Help
2
Replies
44
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
506
  • Introductory Physics Homework Help
Replies
17
Views
413
  • Introductory Physics Homework Help
Replies
2
Views
664
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top