Capacitance of a System (Spherical Conducting Shells)

But this was a simple and basic case.)In summary, the system consists of two concentric spherical conducting shells separated by a dielectric material of dielectric constant 10 εo. Using the equation C=(Q/V)=4πεo(ab/a-b), and taking into account the linearity of the dielectric, the capacitance of the system can be calculated. The presence of the free charges on the conducting shells is not relevant for this calculation.
  • #1
GermanMC
6
0

Homework Statement


A spherical conducting shell of radius 0.1 m has a free charge of -2 μC. It is surrounded by a concentric spherical conducting shell of radius 0.12 m carrying a free charge of -2 μC. Between the shells is a dielectric material of dielectric constant 10 εo. If the dielectric material is linear, what is the capacitance of the system?


Homework Equations


C=(Q/V)=4πεo(ab/a-b)


The Attempt at a Solution



I attempted the solution with only the above equation, I'm just not sure how to to include the dielectric constant in my answer. The instructions say that I should use one of Maxwell's Equations. How do I make the connections I need?
 
Physics news on Phys.org
  • #2
GermanMC said:

Homework Statement


A spherical conducting shell of radius 0.1 m has a free charge of -2 μC. It is surrounded by a concentric spherical conducting shell of radius 0.12 m carrying a free charge of -2 μC. Between the shells is a dielectric material of dielectric constant 10 εo. If the dielectric material is linear, what is the capacitance of the system?


Homework Equations


C=(Q/V)=4πεo(ab/a-b)


The Attempt at a Solution



I attempted the solution with only the above equation, I'm just not sure how to to include the dielectric constant in my answer. The instructions say that I should use one of Maxwell's Equations. How do I make the connections I need?

Nvm, I feel that I have the kinks worked out.
 
  • #3
I was going to say, the first thing is to ignore the bit about the -2μC charges!
 
  • #4
Bingo. Free charge didn't matter.
 
  • #5
GermanMC said:
Bingo. Free charge didn't matter.

It might have if the dielectric were nonlinear, but colleges don't pull stuff like that on you!
(In that case, C would have to be defined as dq/dV instead of q/V, or "incremental capacitance" instead of just "capacitance". Many types of real-life capacitors do display some nonlinearity, matter of fact.
 

What is capacitance and how is it related to spherical conducting shells?

Capacitance is a measure of an object's ability to store electrical charge. In the case of spherical conducting shells, the capacitance is determined by the size and distance between the shells, as well as the dielectric material between them.

What is the formula for calculating the capacitance of a system of spherical conducting shells?

The formula for calculating the capacitance of a system of spherical conducting shells is C = 4πε0εr(r2-r1), where ε0 is the permittivity of free space, εr is the relative permittivity of the dielectric material, and r1 and r2 are the radii of the inner and outer shells, respectively.

How does the capacitance change if the distance between the shells is increased?

If the distance between the shells is increased, the capacitance of the system will decrease. This is because a larger distance between the shells means a smaller electric field and therefore, a smaller ability to store charge.

What is the effect of using a dielectric material between the shells on the capacitance?

Using a dielectric material between the shells increases the capacitance of the system. This is because the dielectric material has a higher relative permittivity than air, which allows for a larger electric field and therefore, a higher ability to store charge.

How does the capacitance of a system of nested spherical shells compare to that of a single spherical shell?

The capacitance of a system of nested spherical shells is equivalent to that of a single spherical shell with a radius equal to the outermost shell. This is because the inner shells shield the outer shells, effectively creating a single conducting surface.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
3K
Replies
4
Views
350
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
29
Views
3K
  • Advanced Physics Homework Help
Replies
6
Views
4K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
959
Back
Top