- #1

rofldude188

- 5

- 0

- 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