Solving Capacitance Problem: Identical Answers Explained

In summary, the two problems involve identical setups with concentric conducting shells and a dielectric material between them. The capacitance between A and C is the same in both cases and can be calculated using the formula (4πεo.Kabc)/(Ka(c-b)+c(b-a)). This is because the presence of the middle conductor B does not change the electric field in the two regions. The potential is constant on a metal surface and the equipotential surfaces are concentric spheres. Even when a thin metal shell is placed at radius b, the setup remains identical with the first one. The electric field may induce charges on the metal surfaces, but the net charge of the shell remains zero. The electric field can be calculated using Gauss'
  • #1
erisedk
374
7

Homework Statement


Why are both these problems identical?
In both cases, c>b>a.
Q1 Three concentric conducting shells A, B and C of radii a, b and c are arranged as shown. A dielectric of dielectric constant K is filled between A and B. Find the capacitance between A and C.

Q2 A spherical capacitor is made of two conducting spherical shells of radii a and c. The space between the shells is filled with a dielectric of dielectric constant K unto a radius b as shown. Find the capacitance of the system.

Homework Equations

The Attempt at a Solution


Both have answers (4πεo.Kabc)/(Ka(c-b)+c(b-a))
I know how to do the first one, but I don't understand why both have the same answer, ie. why the capacitance of the system is equal to the capacitance between A and C.
 
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  • #2
Assume a charge of Q on the inner conductor. Does the presence of the conductor B change the electric field in the two regions between the two cases?
 
  • #3
vela said:
Assume a charge of Q on the inner conductor. Does the presence of the conductor B change the electric field in the two regions between the two cases?
I know it doesn't mathematically, but I'm not getting a feel for it.
 
  • #4
The potential is constant on a metal surface. Because of spherical symmetry, the equipotential surfaces are concentric spheres between A and C.
Replacing an equipotential surface with a very thin metal foil does not change anything, the electric field stays the same. That thin metal shell can be placed at radius b, and nothing is changed, but the set-up becomes identical with the first one.
 
  • #5
ehild said:
The potential is constant on a metal surface. Because of spherical symmetry, the equipotential surfaces are concentric spheres between A and C.
Replacing an equipotential surface with a very thin metal foil does not change anything, the electric field stays the same. That thin metal shell can be placed at radius b, and nothing is changed, but the set-up becomes identical with the first one.
But won't the metal surface induce charges on it's inner and outer surfaces, and thus alter the electric field?
 
  • #6
The electric field induce charge on both surfaces of the metal, but the net charge of the shell remains zero.
Well, calculate the electric field below and above the middle metal shell, applying Gauss' Law.
 
  • #7
Got it! Thanks :)
 

Related to Solving Capacitance Problem: Identical Answers Explained

1. How do I solve identical capacitance problems?

Solving identical capacitance problems involves using the formula C = Q/V, where C is the capacitance, Q is the charge, and V is the voltage. You can also use the formula C = εA/d, where ε is the permittivity of the material, A is the area of the plates, and d is the distance between the plates.

2. What is the difference between series and parallel capacitors?

In series capacitors, the capacitance is reduced and the total voltage is divided among the capacitors. In parallel capacitors, the capacitance is increased and the voltage remains the same across each capacitor.

3. How do I determine the equivalent capacitance in a circuit?

For series capacitors, the equivalent capacitance is found by adding the individual capacitances. For parallel capacitors, the equivalent capacitance is found by adding the reciprocals of each individual capacitance and then taking the reciprocal of the sum.

4. What are some common mistakes to avoid when solving capacitance problems?

Some common mistakes to avoid include forgetting to convert units, using the wrong formula, and not taking into account the dielectric material between the plates. It is also important to double check your calculations and make sure all values are entered correctly.

5. How can I apply capacitance problems to real-life situations?

Capacitance problems can be applied to various real-life situations such as designing electrical circuits, calculating the energy storage capacity of a capacitor, and understanding how capacitors are used in electronic devices. They can also be used to analyze and improve the efficiency of power transmission and distribution systems.

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