# Bound surface charge on a linear dielectric half-cylinder

#### Daniel Sellers

1. The problem statement, all variables and given/known data
Problem statement in attached photo. This is an ungraded assigned problem which I am using to study for an exam, so I don't need the whole solution just help with a couple of points I am confused about.

One: Part d) is really important to how I will answer part b). If we can use Gauss's law to find D then part b) should be very easy. If we can't then I suppose I will have to use boundary values to solve for P somehow in order to find the bound charges.

I don't see any reason why we shouldn't be able to, but one (very flawed) online solution said we could not, and the fact that we are asked the question makes me nervous about proceeding this way, so I wanted to ask some of the critical experts that frequent this forum.

Two: Why is the fact Xe << 1 relevant? This also seems to point me towards using boundary value equations to solve the problem, but I'm simply not familiar enough with this material yet to confidently set up the problem.

2. Relevant equations

P = ε0χeE

Gauss's law for electric displacement?

D = ε0E + P = εE

3. The attempt at a solution

I am fairly confident that I have solved part a) correctly and that I can solve the rest if the above points are clarified

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#### TSny

Homework Helper
Gold Member
1. The problem statement, all variables and given/known data
One: Part d) is really important to how I will answer part b). If we can use Gauss's law to find D then part b) should be very easy. If we can't then I suppose I will have to use boundary values to solve for P somehow in order to find the bound charges.

I don't see any reason why we shouldn't be able to...
In order to use Gauss' law, there needs to be enough symmetry so that there will exist a particular shape of the Gaussian surface that will allow you to simplify $\oint \mathbf{D \cdot da}$. So, think about whether there is sufficient symmetry in this problem.

Two: Why is the fact Xe << 1 relevant?
This will allow you to use approximations in which you only need to be accurate to first order in $\chi_e$. For example, it should be easy to find an expression for the polarization $\mathbf{P}$ that is accurate to first order in $\chi_e$ if you know $\mathbf{E}$ due to the line charge.

Last edited:
• Daniel Sellers

#### Daniel Sellers

When considering the free charge only, we have cylindrical symmetry, but that symmetry does not apply to the entire problem. So for now I will abandon using Guass's law.

As for χe, assuming that it is very small let's us say that ε ≈ ε0. This would suggest that the electric field in the dielectric is approximately the field we would see from the charged wire alone.

Then P ≈ χeλ/(2πr) in the radial direction.

Is this close to what you meant?

#### TSny

Homework Helper
Gold Member
When considering the free charge only, we have cylindrical symmetry, but that symmetry does not apply to the entire problem. So for now I will abandon using Guass's law.
Right.

As for χe, assuming that it is very small let's us say that ε ≈ ε0. This would suggest that the electric field in the dielectric is approximately the field we would see from the charged wire alone.

Then P ≈ χeλ/(2πr) in the radial direction.

Is this close to what you meant?
Yes, exactly what I meant. • Daniel Sellers

#### Daniel Sellers

Terrific! Back to practicing electrostatics problems then.

Thanks for your help!

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