Capicatance of spherical capacitor

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SUMMARY

The capacitance of a spherical capacitor, consisting of a smaller conducting sphere of radius 'a' and a larger conducting shell of radius 'b', is defined by the formula C = 4πε₀ab/(b - a). This formula is derived using Gauss' law and the relationship between charge (Q), capacitance (C), and voltage (V). The discussion highlights that capacitance does not depend on the charges of the conductors but rather on their geometry and grounding conditions. Grounding can affect charge distribution and the effective capacitance of the system.

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  • Understanding of Gauss' law
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  • Knowledge of capacitance and its mathematical definitions
  • Basic principles of electrostatics and charge distribution
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songoku
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Homework Statement


A spherical capacitor consists of a spherical conducting sheell of radius b and charge -Q concentricc with a smaller conducting sphere of radius a and charge +Q. The capacitance of this device :

a. zero

b. \frac{ab}{4\pi\epsilon_0(b-a)}

c. \frac{ab}{4\pi\epsilon_0(a-b)}

d. \frac{4\pi\epsilon_0 ab}{b-a}

e. \frac{4\pi\epsilon_0 ab}{a-b}


Homework Equations


Q = CV

V=k\frac{Q}{r}

The Attempt at a Solution



V_1=k\frac{-Q}{b}

V_2=k\frac{Q}{a}

Then, how to continue?

Thanks
 
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By applying Gauss' law to an charged conducting sphere, the electric field outside it is found to be
E = kQ*r^2.
The potential difference delta V = (kQ/r^2)*dr

The voltage between the spheres can be found by integrating the electric field along a radial line:
V =kQ*intg(1/r^2)*dr between a to b.
From the definition of capacitance, the capacitance is
C = Q/V.
 


Hi rl.bhat

V =kQ*intg(1/r^2)*dr between a to b = kQ (1/a - 1/b). This will be the same as V2 + V1 from my previous work.

Should the potential difference be : V2 - V1 = kQ (1/a + 1/b)?

Thanks
 


Inside the sphere the electric field is due to the smaller sphere only. The potential of the inner sphere with respect to the outer sphere is
Vab = Va - Vb = kQ (1/a - 1/b).
 


Hi rl.bhat

I think I get your point. Now I have new questions.

1.
What if the charges are reversed (the outer sphere is Q and inner sphere is -Q)? Will it be :
V =k(-Q)*intg(1/r^2)*dr between a to b

2.
What if the inner and outer spheres have the same charge, e.g -Q ?Thanks
 


Capacitance of the capacitor does not depends on the charges of the conductors. While performing its work it must have opposite charges on them.
Capacitance of capacitor in the given problem depends on which conductor is grounded.
 


rl.bhat said:
Capacitance of the capacitor does not depends on the charges of the conductors.
so if the charges are reversed (the outer sphere is Q and inner sphere is -Q), it will be the same : V =kQ*intg(1/r^2)*dr between a to b ?

While performing its work it must have opposite charges on them.
then if the inner and outer spheres have the same charge, it is not capacitor, it is just conducting sphere with charge?

Capacitance of capacitor in the given problem depends on which conductor is grounded.
In the question, the conductor grounded is the outer sphere?

Thanks
 


songoku said:
so if the charges are reversed (the outer sphere is Q and inner sphere is -Q), it will be the same : V =kQ*intg(1/r^2)*dr between a to b ?
Yes.[/QUOTE]then if the inner and outer spheres have the same charge, it is not capacitor, it is just conducting sphere with charge?[/QUOTE]

It is not possible to have the same charge on the concentric spherical conductors out of which one is grounded.

[/QUOTE]In the question, the conductor grounded is the outer sphere?[/QUOTE]

In that case the outer sphere cannot be charged by any external source.
Even you can ground the inner surface of the inner sphere instead of outer surface of the outer sphere.
 


rl.bhat said:
It is not possible to have the same charge on the concentric spherical conductors out of which one is grounded.
I don't read any infomation from the question about grounding. The outer sphere is more negative, so does it mean that the outer sphere is grounded?

In that case the outer sphere cannot be charged by any external source.
Even you can ground the inner surface of the inner sphere instead of outer surface of the outer sphere.
If I ground the inner sphere, will it be more negative than the outer sphere?

Thanks
 
  • #10


songoku said:
I don't read any infomation from the question about grounding. The outer sphere is more negative, so does it mean that the outer sphere is grounded?
If you give a positive charge to the inner sphere, a negative charge will induce on the inner surface of the outer sphere and a positive charge on the outer surface of the outer sphere. Since in the problem, it is mentioned that outer sphere has negative charge, it is obvious that its outer surface is grounded. Negative charge on the inner remain there because they are bounded by the positive charge on the inner sphere.
songoku said:
If I ground the inner sphere, will it be more negative than the outer sphere?
Grounding does not increase the charge on any sphere.
If you want to ground the inner sphere, you have to ground the inner surface of the inner sphere. In that case, the system becomes the parallel combination of two capacitors
 
  • #11


Hi rl.bhat

What will happen if I ground the outer surface of the inner sphere and why the system become parallel combination if I ground the inner surface of inner sphere?

Sorry for asking so many questions. Thanks
 
  • #12


songoku said:
What will happen if I ground the outer surface of the inner sphere and why the system become parallel combination if I ground the inner surface of inner sphere?
Then you will have a single spherical capacitor of larger radius. Inner sphere is irrelevant. Charge will be stored only on the outer surface of the outer sphere.
If you ground the inner surface of inner sphere, charge can be stored on inner and outer surfaces of the outer sphere.
The combined capacitance will be
C = 4π*εο*ab/(b - a) + 4π*εο*b
Usually the capacitors are charged by removing the electrons from the conductors by any method. When you remove electrons from the outer surface of the outer sphere, it becomes positive. Electrons from the inner surface will move towards the outer surface making it positive. This positive charge will induce negative charge on the outer surface of the inner sphere and positive charge on the inner surface of the inner sphere. The ground connection removes these positive charges from the inner surface.
 
  • #13


Thanks a lot for your help !
 

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