A dielectric hemisphere

[sentinel]

Homework Statement

we put a dielectric hemishphere with radius (a) and dielectric coeeficient(k) on an infinite conducting plane.the system(hemsiphere and the plane) are in a constant electric field E0.
find the surface charge distrubution on the plane as a function of (r).
and explain what happens?

Homework Equations

E in =1/k E (in a dielectric E weakens)
maxwell equations
e0E=surface charge(charge per area)

The Attempt at a Solution

for the case k=1 there won't be any charge on the plane(I think)
also I tried to think what actually happens?will there be any charge inside the hemisphere?
and these are (I think) the solutions:

 

[sentinel]

anyone??

 

[SammyS]

Homework Statement

we put a dielectric hemisphere with radius (a) and dielectric coefficient(k) on an infinite conducting plane.the system(hemisphere and the plane) are in a constant electric field E0.
find the surface charge distribution on the plane as a function of (r).
and explain what happens?

What coordinate is r ?

Homework Equations

E in =1/k E (in a dielectric E weakens)
Maxwell's equations
e0E=surface charge(charge per area)

The Attempt at a Solution

for the case k=1 there won't be any charge on the plane(I think)

There is a charge on the plane, if the electric field is perpendicular to the plane.

Use Gauss's Law.

also I tried to think what actually happens?will there be any charge inside the hemisphere?
and these are (I think) the solutions:

 

[TSny]

The Attempt at a Solution

for the case k=1 there won't be any charge on the plane(I think)
also I tried to think what actually happens?will there be any charge inside the hemisphere?
and these are (I think) the solutions:

Shouldn't k = 1 correspond to a vacuum (i.e., no hemisphere present)? Then the field should just equal E_o everywhere.

I don't know your background, but you can solve this problem as a boundary value problem for the electric potential. Then use the potential to get the field at the surface of the plane, and hence the charge density.

You can also relate this problem to the problem of a dielectric sphere placed in an external uniform field with no conducting plane.

 

[paranormal]

Your solution is correct for the case when k=1, as there will be no charge on the plane due to the presence of the dielectric hemisphere. This is because the electric field inside the dielectric is weaker than the external field E0, and therefore there is no need for any surface charge on the plane to maintain the electric field.

However, for the case when k≠1, there will be a surface charge density on the plane due to the presence of the dielectric hemisphere. This is because the electric field inside the dielectric is weaker than the external field E0, and therefore there is a need for a surface charge on the plane to maintain the electric field.

To find the surface charge distribution, we can use the boundary conditions at the interface between the dielectric and the conducting plane. These boundary conditions state that the tangential component of the electric field must be continuous across the interface, and the normal component of the electric displacement must be continuous across the interface. Using these boundary conditions, we can find an expression for the surface charge density on the plane as a function of (r).

What happens in this system is that the presence of the dielectric hemisphere modifies the electric field in the region around it. This can be seen by comparing the electric field inside the dielectric (E in = E0/k) to the electric field outside the dielectric (E out = E0). The electric field is weaker inside the dielectric, and this leads to a redistribution of charge on the plane to maintain the electric field. This redistribution of charge results in a non-uniform surface charge distribution on the plane, with a higher charge density near the edges of the hemisphere. This phenomenon is known as polarization, where the presence of a dielectric material in an electric field results in the formation of a dipole moment in the material.