- #1

niko_.97

- 18

- 4

Moved from a technical forum, so homework template missing

This is problem 4.13 from Griffiths (edition 3).

The question asks:

A very long cylinder, of radius a, carries a uniform polarization P perpendicular

to its axis. Find the electric field inside the cylinder. [Careful: I said "uniform," not "radial"!]

I decided to try and find the bound charges. I am not really sure what he means exactly by uniform and perpendicular to the axis, so I just took it as being in the x hat direction. (i.e

The bound volume charge is zero since the divergence of x hat is 0. Then the bound surface charge I got as being σ

However, the solutions do it a different method.

Think of it as two cylinders of opposite uniform charge density ±ρ. Inside, the field at a distance s from

the axis of a uniformly charge cylinder is given by Gauss’s law: E2πsl = (1/ε

two such cylinders, one plus and one minus, the net field (inside) is

the total dipole moment of a chunk of length l is

I think I understand the solution. although why d=/=0 isn't exactly clear to me.

But my main query is why my solution gives you a different answer?

The question asks:

A very long cylinder, of radius a, carries a uniform polarization P perpendicular

to its axis. Find the electric field inside the cylinder. [Careful: I said "uniform," not "radial"!]

I decided to try and find the bound charges. I am not really sure what he means exactly by uniform and perpendicular to the axis, so I just took it as being in the x hat direction. (i.e

**P**=P**x̄)**The bound volume charge is zero since the divergence of x hat is 0. Then the bound surface charge I got as being σ

_{b}= Pcos(φ) for 0≤φ≤π and σ_{b}= -Pcos(φ) for π<φ<2π for the curved side of the cylinder and then 0 for both the top and bottom face. If you integrate this to find the potential you get zero (since the integral of cosφ from 0 to π is 0). Obviously then the electric field is also 0.However, the solutions do it a different method.

Think of it as two cylinders of opposite uniform charge density ±ρ. Inside, the field at a distance s from

the axis of a uniformly charge cylinder is given by Gauss’s law: E2πsl = (1/ε

_{0})ρπs^{2}l =>**E**= (ρ/2ε_{0})**s**. Fortwo such cylinders, one plus and one minus, the net field (inside) is

**E**=**E**+_{+}**E**= (ρ/2ε_{-}_{0}) (**s**−_{+}**s**). But_{−}**s**−_{+}**s**= −_{−}**d**, so**E**= −ρ**d**/(2ε_{0}), where**d**is the vector from the negative axis to positive axis. In this casethe total dipole moment of a chunk of length l is

**P**πa^{2}l=ρπa^{2}l**d**. So => ρ**d**=**P**, and E =−**P**/(2ε_{0}), for s < a.I think I understand the solution. although why d=/=0 isn't exactly clear to me.

But my main query is why my solution gives you a different answer?