Solving Dielectric Problem: Electric Field & Charge Density

• rofldude188
In summary, the electric field in space for r < R is 0 because the gold ball is a conductor. For r > R, the electric field is given by E = Q / (4 * pi * r^2 * epsilon_o * (1 + epsilon_r)), where Q is the charge released on the ball and epsilon_o and epsilon_r are the permittivities of air and oil respectively. The factor of 1/2 in the answer is due to the two dielectrics being in "parallel". As for the charge density, the correct values for the top and bottom sides of the ball are rho_s = Q / (2 * pi * (1 + epsilon_r) * R^2) and rho_s =
rofldude188
Homework Statement
A solid gold ball is floated on an oil bath so that half the ball is above the oil and the other half is immersed in oil. The oil has a relative permittivity of εr. Charge Q is released on the ball. Calculate the electric field in space. What is the charge density on the ball for the part above the oil and in the oil.
Relevant Equations
surface integral D * dS = Q_free
D = E * epsilon_o

a) Find the electric field in space
For r < R (where R is the radius of the gold ball), E = 0 because the gold ball is a conductor.
For r > R, let us make a Gaussian surface. $$\int \vec{D} \cdot \vec{dS} = Q_{free} \implies \vec{D} = \frac{Q}{4 \pi r^2}$$
Now this is a bit hand wavy but the two dielectrics (air and oil) appear to be in "parallel" so we can just add them together as such $$\epsilon _{total} = \epsilon_o + \epsilon_o \epsilon_r$$
$$\implies \vec{E} = \frac{Q}{4 \pi r^2 \epsilon_o (1 + \epsilon_r)}$$
The problem is the answer is $$\vec{E} = \frac{Q}{2 \pi r^2 \epsilon_o (1 + \epsilon_r)}$$ where is this factor of 1/2 coming from? b) As for charge density, I'm not sure how to calculate this. The answers are though $$Top side: \rho_s = \frac{Q}{2 \pi (1 + \epsilon_r) R^2}$$ $$Bottom side: \rho_s = \frac{\epsilon_r Q}{2 \pi (1 + \epsilon_r) R^2}$$

Any help would be appreciated

rofldude188 said:
two dielectrics (air and oil) appear to be in "parallel" so we can just add them together as such
So if it were air above and below you would use 2ε0?

Btw, this must be very dense oil.

rofldude188 said:
Problem Statement: A solid gold ball is floated on an oil bath so that half the ball is above the oil and the other half is immersed in oil. The oil has a relative permittivity of εr. Charge Q is released on the ball. Calculate the electric field in space. What is the charge density on the ball for the part above the oil and in the oil.
Relevant Equations: surface integral D * dS = Q_free
D = E * epsilon_o

View attachment 245923

a) Find the electric field in space
For r < R (where R is the radius of the gold ball), E = 0 because the gold ball is a conductor.
For r > R, let us make a Gaussian surface. $$\int \vec{D} \cdot \vec{dS} = Q_{free} \implies \vec{D} = \frac{Q}{4 \pi r^2}$$
Now this is a bit hand wavy but the two dielectrics (air and oil) appear to be in "parallel" so we can just add them together as such $$\epsilon _{total} = \epsilon_o + \epsilon_o \epsilon_r$$
$$\implies \vec{E} = \frac{Q}{4 \pi r^2 \epsilon_o (1 + \epsilon_r)}$$
The problem is the answer is $$\vec{E} = \frac{Q}{2 \pi r^2 \epsilon_o (1 + \epsilon_r)}$$ where is this factor of 1/2 coming from?b) As for charge density, I'm not sure how to calculate this. The answers are though $$Top side: \rho_s = \frac{Q}{2 \pi (1 + \epsilon_r) R^2}$$ $$Bottom side: \rho_s = \frac{\epsilon_r Q}{2 \pi (1 + \epsilon_r) R^2}$$

Any help would be appreciated
Question to yourself: What is the relative charge distribution above and below the oil-air interface?

Hint: compute the potential of the ball as integrated E field in two paths: in air and thru oil. Remember the potential of the ball is the same thruout the ball. What does that tell you about the two E fields integrated to infinity?
Then, what needs to happen at the oil-air interface to satisfy this requirement?

Note that your answer for the charge densities could not be correct since total charge must = Q.

1. What is a dielectric material?

A dielectric material is an insulating material that does not conduct electricity. It is commonly used in electronic devices to prevent the flow of electric current and to store electrical energy.

2. How does a dielectric material affect the electric field?

Dielectric materials have a property called permittivity, which describes how easily they can become polarized when placed in an electric field. This polarization creates an opposing electric field, reducing the overall strength of the original electric field.

3. What is the relationship between electric field and charge density in a dielectric material?

In a dielectric material, the electric field is directly proportional to the charge density. This means that as the electric field increases, so does the charge density.

4. How do you solve for the electric field and charge density in a dielectric problem?

To solve a dielectric problem, you must first determine the permittivity of the material and the boundary conditions of the system. Then, you can use the equations for electric field and charge density to solve for their values at different points within the material.

5. Can dielectric materials be used to enhance the strength of an electric field?

Yes, dielectric materials can be used in capacitors to increase the overall strength of an electric field. This is because the polarization caused by the dielectric material decreases the electric field between the capacitor's plates, allowing for a larger potential difference to be maintained.

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