Spherical capacitor (Irodov 3.101.)

[Biotic]

Homework Statement

Find the capacitance of an isolated ball-shaped conductor of radius R₁ sorrounded by an adjacent concentric layer of dielectric with permitivity ε and outside radius R₂.

Homework Equations

The Attempt at a Solution

The official solution says something like:

Difference of potentials is --- int(R1,R2)(E*dr) + int(R2,∞)(E*dr)

Sorry, but I don't know how to write equations nicely.
I don't quite understand what is the second integral here for. Can someone explain?

 

[Dickfore]

The electric field only exists between the two conductors, and they carry equal by magnitude, but opposite by sign charges $Q$, and $-Q$, respectively. Using Gauss's Law, you should get:
$$E(r) = \frac{Q}{4 \pi \epsilon} \, \frac{1}{r^2}$$
and the direction is along the radius.

The potential difference is:
$$V_1 - V_2 = \int_{R_1}^{R_2}{E(r) \, dr} = \frac{Q}{4 \pi \epsilon} \, \int_{R_1}^{R_2}{\frac{dr}{r^2}}$$

To perform this integral, I suggest you use the following subsitution:
$$r = A \, x^\alpha \Rightarrow dr = A \, \alpha \, x^{\alpha - 1} \, dx$$
Then, the integrand becomes:
$$\frac{A \, \alpha \, x^{\alpha - 1} \, dx}{A^2 \, x^{2\alpha}} = \frac{\alpha}{A} \, x^{-1 - \alpha} \, dx$$
What would be the most convenient choice for $\alpha$? Does it depend what you choose for $A$ (look at how the bounds of integration are changing)?

 

[Biotic]

Thank you for the reply.

I know how to solve integrals. What I don't know is... Why are there two integrals in the official solution for potential difference? One from R1 to R2 and the other from R2 to infinity. If you don't have it, the soultions for irodov are available online so if you could check it out and tell me what the second integral is there for, that would be great.

 

[Dickfore]

I haven't seen the solution online. If you know the link, post it here.

There is an integral from $R_2$ to $\infty$ because the electric field is not zero, since there is no exterior grounded spherical shell. I misread the formulation of the problem.

By the way, you should use Gauss's Law for the dispalcement vector $\vec{D}$, which is

Flux of D = total enclosed free charge

and then find $E = D/\epsilon$ in the corresponding region. You can see that the electric field is discontinuos at the boundary.

Effectively, you have two spherical capacitors connected in series.

 

[asteriatic]

I would like to provide a more detailed explanation and solution to the problem.

First, let's define some terms and equations that are relevant to this problem:

- Capacitance: The ability of a system to store an electric charge, measured in Farads (F).
- Spherical capacitor: A type of capacitor where the two conductors are spherical in shape and are separated by a dielectric material.
- Dielectric: An insulating material that can be placed between two conductors to increase the capacitance of a system.
- Permittivity (ε): A measure of a material's ability to store electrical energy in an electric field.
- Radius (R): The distance from the center of a circle or sphere to its outer edge.
- Electric field (E): A vector field that describes the strength and direction of the electric force at any given point.

Now, let's look at the problem at hand. We have a spherical conductor with radius R1, surrounded by a concentric layer of dielectric with permitivity ε and outside radius R2. We want to find the capacitance of this system.

To solve this problem, we can use the equation for capacitance of a spherical capacitor, which is given by:

C = 4πεR1R2 / (R2 - R1)

This equation takes into account the radius of the inner conductor (R1), the radius of the outer conductor (R2), and the permittivity of the dielectric material (ε).

Now, let's break down the solution provided in the problem. The first integral, int(R1,R2)(E*dr), represents the electric field between the two conductors (from R1 to R2). This integral is used to calculate the potential difference between the two conductors, which is a key factor in determining capacitance.

The second integral, int(R2,∞)(E*dr), represents the electric field outside of the outer conductor (from R2 to infinity). This integral takes into account the effect of the dielectric material on the electric field. As the electric field passes through the dielectric, it is weakened, which affects the potential difference between the two conductors.

In summary, the second integral is necessary to accurately calculate the potential difference between the two conductors, and therefore, the capacitance of the system. Without taking into account the effect of the dielectric material, the calculated capacitance would be incorrect.