# Wire surrounded by a linear dielectric in a uniform E field

• physicsisfun0
In summary, we have an uncharged, conducting wire surrounded by a linear dielectric material and placed in an external electric field. The electric potential inside (a < s < b) can be represented by Vinbetween=Acosφ + (B/s)cosφ, and outside (s > b) by Voutside=-Eoscosφ + (D/s) cosφ. The boundary conditions are set to ensure continuity of the electric potential and electric field. The correct application of these conditions will allow for the determination of the surface density on the dielectric of a bound and free charge at s=a and s=b. The current attempt at a solution may not be accurate due to potential errors in applying the boundary conditions and dimensionally inconsistent terms
physicsisfun0

## Homework Statement

We have an uncharged, conducting wire with radius a. We surround it by a linear dielectric material, εr, which goes out to radius b. We place this in an external electric field, Eo.

## Homework Equations

We have electric potential inside (a < s < b)
Vinbetween=Acosφ + (B/s)cosφ
and outside (s > b)
Voutside=-Eoscosφ + (D/s) cosφ

Our boundary conditions are:
Vinbetween=Voutside
εrEsinrEsout
when s = a, V = 0
when s = b, V = -Eoscosφ

## The Attempt at a Solution

I have solved for the constant and got:
A = - B/a2
B = (-Eob2 + D)/(b2(a-2 + a-2)
D = -((εrb2Eo + εrBEo + (Eo/a2) + (Eo/a2))b)/(b(a-2 + b-2 - εr + εrb)

Is this correct? Once I know these, I can find electric potential and electric field. From there, how would I go about solving for surface density on the dielectric of a bound and free charge, when s=a and s=b.

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It is not clear that you have applied the boundary conditions correctly. Presumably there is vacuum for s > b therefore there ought to be terms with ε0 coming from the continuity of the normal component of ##\vec D## across the boundary, $$\epsilon_r \frac{\partial V_I}{\partial s}=\epsilon_0 \frac{\partial V_{II}}{\partial s}$$ Also the sum in the expression for coefficient D (Eo/a2))b)/(b(a-2 + b-2 - εr + εrb) is dimensionally inconsistent.

## 1. What is a linear dielectric?

A linear dielectric is a non-conductive material that can be polarized by an electric field. This means that the material's atoms or molecules will align in response to the electric field, resulting in an overall charge separation within the material.

## 2. How does a wire surrounded by a linear dielectric behave in a uniform E field?

In a uniform electric field, the wire surrounded by a linear dielectric will experience a force due to the electric field. This force will cause the wire and the dielectric material to polarize, resulting in a net charge on the wire and the dielectric.

## 3. What is the purpose of using a linear dielectric in this scenario?

The presence of a linear dielectric can increase the capacitance of the wire in the electric field. This is because the dielectric material increases the effective distance between the charges on the wire, resulting in a larger potential difference for the same amount of charge.

## 4. How does the dielectric constant of the material affect the behavior of the wire in the electric field?

The dielectric constant, also known as the relative permittivity, determines the degree to which a material can be polarized. A higher dielectric constant means that the material can be more easily polarized, resulting in a larger capacitance and a stronger force on the wire in the electric field.

## 5. Is the behavior of the wire surrounded by a linear dielectric in a uniform E field affected by the shape or size of the wire?

Yes, the shape and size of the wire can affect its behavior in the electric field. A longer wire will experience a greater force, while a thicker wire will have a larger capacitance. Additionally, the shape of the wire can affect how the electric field interacts with the wire and the surrounding dielectric material.

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