# Electric field with dielectric

• chrishans
In summary: Yes, it should be fine to calculate it treating it as free charges, because your density does include polarization charges.
chrishans

## Homework Statement

There's a ring of radius R with linear uniform charge density $$\lambda=constant$$ (which includes both free charges and polarization charges). The ring is in a dielectric region with $$\varepsilon_r$$. I have to find the electric field on any point of the axis of the ring

## Homework Equations

$$\vec E=\int \frac{\lambda dl (\vec r - \vec r')}{| \vec r - \vec r' | ^3}$$

## The Attempt at a Solution

I actually don't know and coulnd't find anywhere if I can calculate the electric field using the above formula when $$\lambda$$ includes free AND polarization charges.

Is there a coefficient missing in your equation ?

No, that's all I have. Since I'm given a linear density for the ring which includes free and polarization charges, I assume it's a dielectric, but I don't know its coefficient. With that density which includes polarization charges, is it ok to calculate the electric field, like in the formula I wrote before?

Since the ring is submerged in a dielectric region with εr, the Electric field decreases by a factor of εr.
So E' = E/εr, where E is the magnitude of the field without the dielectric, and E' with dielectric.
Now take an infinitely small section on the circumference on the ring, dl. The charge associated is dq=λdl.
The field associated with dq is 1/4piε (dq/r^2). However, since the ring is symmetric, the x-component of the net electric field, which is perpendicular to the axis, cancels out. Now all you need to consider is the y component. The y component of the field is Ey= E'cos(theta), where theta is the angle made between the axis, and the field vector coming from dq. Set "z" as the distance along the axis from the center of the circle. You can find cos(theta) using a relationship between "z" and "R", with the use of pythagorean theorem. Also, when you intergrate with respect to dl, change dl to Rd(theta). Integrate all contributions of dq from 0 to 2pi (because it is a circle) and you should be able to find the electric field anywhere along the axis. I understand that this is a sloppy description, but I hope this help anyways.

It should be fine to calculate it treating it as free charges, because your density does include polarization charges.

chrishans said:
No, that's all I have. Since I'm given a linear density for the ring which includes free and polarization charges, I assume it's a dielectric, but I don't know its coefficient. With that density which includes polarization charges, is it ok to calculate the electric field, like in the formula I wrote before?
Strange. i seem to remember that
$$\vec E= {1\over 4\pi\epsilon_0}\int \frac{\lambda dl (\vec r - \vec r')}{| \vec r - \vec r' | ^3}$$in vacuum. In a medium it would be ##{1\over 4\pi\epsilon}## with ##\epsilon=\epsilon_0 \epsilon_r##, but I could be mistaken. So check.
There is nothing in ##\lambda## that has to do with stuff outside the ring. No change in ##\lambda##, whatever ##\epsilon_r##.

## 1. What is an electric field with dielectric?

An electric field with dielectric refers to the presence of a dielectric material, such as an insulator, in an electric field. This material has a polarizing effect on the electric field, causing it to become weaker compared to the same field in a vacuum.

## 2. How does a dielectric affect the strength of an electric field?

A dielectric material reduces the strength of an electric field by polarizing its molecules in the presence of the field. This causes the electric field to become weaker, resulting in a lower electric potential and electric flux.

## 3. What is the role of a dielectric in capacitors?

In capacitors, a dielectric material is placed between the plates to increase the capacitance. The polarizing effect of the dielectric reduces the electric field between the plates, allowing for a greater amount of charge to be stored on the plates without causing breakdown.

## 4. How do you calculate the electric field with dielectric?

The electric field with dielectric can be calculated using the formula E = E₀/κ, where E₀ is the electric field in a vacuum and κ is the dielectric constant of the material. Alternatively, it can be calculated using the formula E = Q/ε₀A, where Q is the charge on the capacitor plates, ε₀ is the permittivity of free space, and A is the area of the plates.

## 5. What is the significance of dielectric strength in electric fields?

Dielectric strength refers to the maximum electric field that a material can withstand before experiencing breakdown. This is an important factor to consider when designing electrical systems, as exceeding the dielectric strength of a material can lead to failure and potentially dangerous situations.

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