Capacitance ( getting started)

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In summary: The Attempt at a SolutionI believe that I should begin by calculating the area of the insulating spheres with constants k1 and k2 and plug into the equation? or should I treat it as 1 spherical capacitor inside another? this gets 2 equations 4\pi\xi(r(sub b)r(sub a))/(r(sub b) - r(sub a)) and 4\pi\xi(r(sub d)r(sub c))/(r(sub d) - r(sub c)) and treat that as 1 big capacitor?in this way i get 4\pi\xi(r(sub
  • #1
yikes_physics
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Homework Statement



A spherical capacitor is made of two insulating spherical shells with different dielectric constants k1 and k2 situated between two spherical metallic shells and separated by a vacuum gap. Geometrical dimensions of the cross-section are as shown in figure 2. Calculate the capacitance of this system.

Homework Equations


C= Q/V(subscript ab) (couldnt figure out the formatting for subscripts :(


The Attempt at a Solution


I honestly have no idea on how to do this, so that is why i am not looking for help getting the answer, i would just like some help on where to start. Problems without numbers really throw me through a loop and i can usually get them when i get a little jump start.

i believe that i should begin by calculating the area of the insulating spheres with constants k1 and k2 and plug into the equation? or should i treat it as 1 spherical capacitor inside another? this gets 2 equations 4[tex]\pi\xi[/tex](r(sub b)r(sub a))/(r(sub b) - r(sub a)) and 4[tex]\pi\xi[/tex](r(sub d)r(sub c))/(r(sub d) - r(sub c)) and treat that as 1 big capacitor?

in this way i get 4[tex]\pi\xi[/tex](r(sub d)r(sub b))/(r(sub d) - r(sub b))...does this sound right or am i off in the wrong direction?
 

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  • #2
What are: 4(r(sub b)r(sub a))/(r(sub b) - r(sub a)) and 4(r(sub d)r(sub c))/(r(sub d) - r(sub c)) ?

Let the charge on the inside sphere be Q... assume Q>0. let the charge on the outside sphere be -Q.

calculate the voltage between the 2 spheres... first get the field using Gauss' law (you'll have two parts to get the field within each of the dielectrics)... then the integral of -E.dr going from the inner sphere to the outer... that gives the voltage between the 2.

Then C = |Q/V|.
 
  • #3
4(r(sub b)r(sub a))/(r(sub b) - r(sub a)) and 4(r(sub d)r(sub c))/(r(sub d) - r(sub c)) are the 2 little spherical capacitors made up of spheres with radii a and b for the first one and spheres with radii c and d for the second one. no numbers are given, i have to solve C in terms of the radii and the other variables...which is why i don't like these questions, lol.

ill give it another shot, and i appreciate your help in this matter.

take care
 
  • #4
yikes_physics said:
4(r(sub b)r(sub a))/(r(sub b) - r(sub a)) and 4(r(sub d)r(sub c))/(r(sub d) - r(sub c)) are the 2 little spherical capacitors made up of spheres with radii a and b for the first one and spheres with radii c and d for the second one. no numbers are given, i have to solve C in terms of the radii and the other variables...which is why i don't like these questions, lol.

ill give it another shot, and i appreciate your help in this matter.

take care

I forgot about the vacuum gap... so you have 3 sections to get the field for instead of 2 like I initially thought...
 

Related to Capacitance ( getting started)

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in farads (F) and is represented by the symbol C. It is determined by the geometry and materials of the system.

2. What are the factors that affect capacitance?

The factors that affect capacitance include the distance between the plates, the area of the plates, and the dielectric material between the plates. These factors determine the strength of the electric field and the amount of charge that can be stored.

3. How is capacitance calculated?

The capacitance of a system can be calculated using the formula C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage applied. It can also be calculated using the formula C = εA/d, where ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

4. What is the difference between capacitance and capacitance per unit length?

Capacitance per unit length, also known as specific capacitance, is the capacitance of a system per unit length. It is used to describe transmission lines and is measured in farads per meter (F/m). Capacitance, on the other hand, is the overall capacitance of a system and is measured in farads (F).

5. How is capacitance used in practical applications?

Capacitance is used in a variety of practical applications such as in electronic circuits, power factor correction, and energy storage devices like capacitors and batteries. It is also used in touch screens, sensors, and communication systems.

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