Capacitor made of concentric spherical conducting shells

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SUMMARY

The discussion focuses on deriving the capacitance of a capacitor formed by two concentric spherical conducting shells with radii r1 and r3, filled with dielectrics of relative permittivity εr1 and εr2. The capacitance can be calculated by treating the dielectrics as capacitors in series, where the individual capacitances are determined using the formula C = εA/d. The potential difference between the shells is calculated by integrating the electric field, which is reduced by the presence of the dielectrics. The final capacitance expression is derived from the relationship C = Q/V.

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  • Familiarity with integration techniques in physics
  • Concept of capacitors in series and their equivalent capacitance
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perfectionist
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Homework Statement



A capacitor is formed of two concentric spherical conducting shells of radii r1 and r3. The space between the shells is filled from r1 to r2 with a dielectric of relative permittivity [tex]\epsilon[/tex]r1 and from r2 to r3 with a dielectric of relative permittivity [tex]\epsilon[/tex]r2. Derive an expression for the capacitance of this capacitor.

Homework Equations



C = Q/V

The Attempt at a Solution



I know that the dielectric reduces the E. But i seriously do not know where to start from.
 
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perfectionist said:

Homework Statement



A capacitor is formed of two concentric spherical conducting shells of radii r1 and r3. The space between the shells is filled from r1 to r2 with a dielectric of relative permittivity [tex]\epsilon[/tex]r1 and from r2 to r3 with a dielectric of relative permittivity [tex]\epsilon[/tex]r2. Derive an expression for the capacitance of this capacitor.

Homework Equations



C = Q/V

The Attempt at a Solution



I know that the dielectric reduces the E. But i seriously do not know where to start from.





i think that this appears to be a capacitor in series, each with separation distance of
r1-r2 and r2-r3...

using C= epsilon*A/d you can get the individual capacitance and then add it in series...
(where d is the separation distance of each capacitor)

so C = 1/[(1/C1)+(1/C2)]
 
Hi there,

You could also do it in the following way:

Place a charge +Q on the inner conducting shell and a charge -Q on the outer shell. Your challenge is to calculate the potential between the shells as a function of the radial distance from the center of the concentric shells. (Remember to always integrate from the negative plate to the positive plate when calculating potentials from electric fields between capacitor plates)

As you rightly pointed out, the dielectrics reduce the electric field in the regions they occupy. You have been given the dielectric constants and will need them to calculate the electric field between the shells. (What is the relationship between the electric field in vacuum and the electric field in the region of a dielectric with dielectric constant epsilon_sub_r?).

Once you've got the electric field between the shells, you can calculate the potential and hence the capacitance of the configuration using the equation C = Q / V.

Hope this helps.
 

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