Potenital of a cylindrical, linear dialectric in a E-Field

In summary, a long cylinder of linear dielectric material in a uniform electric field results in an equation with a summation over n, where the coefficients of the trigonometric functions must match on each side, and a boundary condition on V must also be taken into account.
  • #1
PhotonSSBM
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Homework Statement


A very long cylinder of linear dialectric material is placed in an otherwise uniform electric field ##E_0##. Find the resulting field within the cylinder. (The radius is ##a##, the susceptibility ##X_e##, and the axis is perpendicular to ##E_0##)

Homework Equations


Boundary conditions
##V_i = V_o## at ##s=a##

##\epsilon \frac{\partial V_i}{\partial n} = \epsilon_0 \frac {\partial V_o}{\partial n}## because there is no free charge

##V = -E_0*s*cos(\theta)## for ##r>>a##

The Attempt at a Solution


I've gotten this far

##\epsilon_r[-na^{-n-1}* (Acos(n*\theta) + Bsin(n*\theta))] = na^{n-1} * (Ccos(n*\theta) + Dsin(n*\theta)) -E_0cos(\theta)##

where n is what I'm summing over. I'm unsure of how to restrict the constants of the trig functions to reduce the problem.
 
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  • #2
This equation isn't correct, as written. There should be a summation over ##n## on each side (except for the term ##-E_0 \cos\theta ~##, which is not summed over ##n##).

For each value of ##n##, the coefficient of ##\cos(n \theta)## on the left side must match the coefficient of ##\cos(n \theta)## on the right side. Similarly for ##\sin(n \theta)##.

So, what condition on the coefficients do you get when matching ##\cos(n \theta)## for the special case of ##n = 1##? What condition on the coefficients do you get when matching ##\cos(n \theta)## for ##n \neq 1##?

Similarly, get equations from matching ##\sin (n\theta)##.

There is also another boundary condition on ##V## (which you stated in the relevant equations section) that you need to invoke where you can again match coefficients on each side.
 
Last edited:

What is the potential of a cylindrical, linear dialectric in an E-Field?

The potential of a cylindrical, linear dialectric in an E-Field is the measure of the work required to move a unit charge from one point to another in the dialectric.

How is the potential of a cylindrical, linear dialectric in an E-Field calculated?

The potential is calculated using the equation V = -E * d, where V is the potential, E is the electric field, and d is the distance between the two points.

What is the relationship between potential and electric field in a cylindrical, linear dialectric?

The relationship between potential and electric field in a cylindrical, linear dialectric is that the electric field is equal to the negative gradient of the potential. In other words, the electric field points in the direction of decreasing potential.

How does the potential of a cylindrical, linear dialectric change with distance?

The potential of a cylindrical, linear dialectric decreases with distance, as the electric field decreases with distance. This means that the potential is highest at the point nearest to the charge and decreases as you move further away.

Can the potential of a cylindrical, linear dialectric ever be negative?

Yes, the potential of a cylindrical, linear dialectric can be negative. This occurs when the electric field is pointing in the opposite direction of the potential gradient, resulting in a negative value for the potential.

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