So, Lightning Decides to Strikes a Sphere

[mars3554]

Homework Statement

1. Lightning strikes a dielectric sphere (Relative Permeability=1.2, Conductivity=10 S/m) of radius 0.1 m at time t=0, depositing uniformly in the sphere a total charge 1 mC. Determine for all t,
the electric field intensity - E(Volts/m) & current density - J(Ampere/m^2) both
inside and outside the sphere.

Homework Equations

Ohm's Law: J = 10 S/m(conductivity) * E
[So if I know one then I know the other]

Electric Field @ Distance R
E = (q\widehat{R}) / (4 \pi \epsilon R ^{2})

The Attempt at a Solution

Don't know where to begin on this one. Almost all of my equations have to do with a dielectric object sitting inside of an E field. Does the charge go dissipate over time? If so then it is asking for an equation correct? Could this be modeled as a point charge in some way?

I'm not looking for an answer only a direction please.

 

[Redbelly98]

Welcome to PF

The charge is initially distributed uniformly throughout the sphere's volume.

The sphere is isolated (apart from the initial lightning strike), so the charge does not go away. However, it will redistribute itself -- i.e. not be uniformly distributed within the volume.

Any thoughts on how/where the charge ends up as t→∞?

 

[mars3554]

Welcome to PF

The charge is initially distributed uniformly throughout the sphere's volume.

The sphere is isolated (apart from the initial lightning strike), so the charge does not go away. However, it will redistribute itself -- i.e. not be uniformly distributed within the volume.

Any thoughts on how/where the charge ends up as t→∞?

Thanks for the reply.

If it as first uniformly distributed, then it will only be that way at t=0 correct?
since it will redistribute afterwards, the answer will be a piecewise function?

I understand that a dielectric in a field causes a dipole, But I'm having trouble visualizing how a dielectric can store charge much less where it will store it. I was under the impression that the material is already "full" but the electrons will spend more time in one direction or are than another causing a moment to occur.

 

[Redbelly98]

Thanks for the reply.

If it as first uniformly distributed, then it will only be that way at t=0 correct?
since it will redistribute afterwards, the answer will be a piecewise function?

Yes, and we are interested in the t≥0 part of the function.

I understand that a dielectric in a field causes a dipole, But I'm having trouble visualizing how a dielectric can store charge much less where it will store it. I was under the impression that the material is already "full" but the electrons will spend more time in one direction or are than another causing a moment to occur.

The dielectric charge stays within the dielectric, but is "redistributed" in the presence of an E-field. Are you familiar with the distinction between bound and free charge?

However, I don't think you need to worry about the bound dielectric charge, at least not right away, because in this equation ...

Electric Field @ Distance R
E = (q\widehat{R}) / (4 \pi \epsilon R ^{2})

... q is referring to the free charge only. Moreover, it's the free charge that lies within a sphere of radius R.