Solving Dielectric Problem: Electric Field & Charge Density

[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

 

[haruspex]

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.

 

[rude man]

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.