Calculating Electric Fields in a Polarized Spherical Shell

In summary, the conversation discusses finding the electric field in a thick, spherical shell made of dielectric material using two methods: using Gauss' Law and using the relation between electric displacement \textbf{D} and electric field \textbf{E}. The correct solution for \textbf{D} is \textbf{P}, and using this value for \textbf{D} will give the same result as using Gauss' Law.
  • #1
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Homework Statement


A thick, spherical shell (inner radius a, outer radius b) is made of dielectric material with a 'frozen in' polarization
[tex]\textbf{P} = \frac{k}{r} \textbf{\hat r} [/tex]
where k is a constant and r the distance from the centre. (There is no free charge). Find the electric field in each o the three regions by two methods:

a) Locate all the bound charge, nd use Gauss' Law to calculate the field it produces
b) Use [tex]\oint \textbf{D}.d\textbf{a} = Q_{f enc} [/tex] to find [tex]\textbf{D} [/tex] and then find the electric field

The Attempt at a Solution



a) bound charge density is [tex]\rho _b -\nabla \textbf{P} = -frac{k}{r^2} [/tex]

Gauss' Law: [tex] \oint E.da = Q_{enc}/\epsilon _0 [/tex]
In the dielectric material:
[tex]Q_{enc} = \int \rho _b = -4\pi \int _a^r dr '= 4\pi k(a-r)[/tex]
I integrated from a to r because that's the only region with charge.
From Guass' Law:

[tex]E4\pi r^2 = 4\pi k(a-r)/\epsilon_0 [/tex]

[tex]E = k(a-r)/(\epsilon_0 r^2)[/tex]

For r<a the electric field is zero because there is no bound charge. Similarly, outside the sphere
[tex]E = k(a-b)/(\epsilon_0 r^2)[/tex]

b)
[tex]\oint \textbf{D}.d\textbf{a} = Q_{f enc} = 0 [/tex]
because there is no free charge. So
[tex] \textbf{D} = \epsilon _0 \textbf{E} + \textbf{P} = 0 [/tex]

[tex] \textbf{E} = -\textbf{P}/\epsilon_0 [/tex]

Where did I go wrong here?
 
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  • #2


Your solution for the electric field using Gauss' Law is correct. However, in part b), you have made a mistake in your calculation for \textbf{D}. Since there is no free charge, the electric displacement \textbf{D} is equal to the polarization \textbf{P}. Therefore, your expression for \textbf{D} should be \textbf{D} = \textbf{P} = \frac{k}{r} \textbf{\hat r}.

Using this value for \textbf{D}, you can then use the relation \textbf{D} = \epsilon_0 \textbf{E} to find the electric field in each of the three regions. This will give you the same result as using Gauss' Law.

Overall, your approach is correct, but you just made a small mistake in your calculation for \textbf{D}. Keep up the good work!
 

FAQ: Calculating Electric Fields in a Polarized Spherical Shell

What is an electric field?

An electric field is a region in space in which an electric charge experiences a force. It is represented by a vector quantity that describes the strength and direction of the force.

How is an electric field in matter created?

An electric field in matter is created by the presence of charged particles, such as protons and electrons, within the material. These charged particles interact with each other and with external electric fields to produce an overall electric field in the material.

What factors affect the strength of an electric field in matter?

The strength of an electric field in matter is affected by the amount and distribution of charged particles in the material, as well as the distance between these particles. The nature of the material, such as its dielectric constant, also plays a role in determining the strength of the electric field.

How is the strength of an electric field in matter measured?

The strength of an electric field in matter is measured using a unit called volts per meter (V/m). This unit represents the amount of force in volts that is exerted on a charge of one coulomb placed at a distance of one meter from the source of the electric field.

What are some practical applications of electric fields in matter?

Electric fields in matter have many practical applications, including in electronics, where they are used to control the flow of electricity in circuits. They are also used in technologies such as capacitors and electric motors. In addition, electric fields in matter play a crucial role in understanding and manipulating the behavior of materials in areas such as material science and engineering.

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