# Electric Field in Cylindrical Insulator

• felguk
In summary, the problem involves finding the electric field inside and outside a solid cylinder with uniform charge density. Using Gauss's law and choosing an appropriate Gaussian surface, the electric field can be calculated by finding the total charge enclosed within the surface. For inside the cylinder, the electric field is equal to p/(2*pi*epsilon_0)*r, while for outside the cylinder it is equal to p*R_0^2/(r^2*epsilon_0).
felguk

## Homework Statement

We have a solid cylinder of radius R0, and it has a uniform charge density p. The length is much greater than its radius, so it appears almost infinite.

## Homework Equations

Find the electric field inside the cylinder where r<R0 and outside the cylinder where r>R0

## The Attempt at a Solution

I know that you need to use Gauss's law on this and I believe that for inside the cylinder it will be E= p/(pi*r^2*ε0)

Will it be the same for outside the insulator?

Hello felguk,

Welcome to Physics Forums!
felguk said:

## The Attempt at a Solution

I know that you need to use Gauss's law on this
Yes, that's right you do need to use Guass' law.
and I believe that for inside the cylinder it will be E= p/(pi*r^2*ε0)
Not quite.

Let's take a look at Guass' law.

$$\oint_S \vec E \cdot d \vec A = \frac{Q_{enc}}{\varepsilon_0}$$

Choose your Gaussian surface such that the electric field's magnitude is constant over the Gaussian surface, and perpendicular to the surface (parallel to the surface's normal vector). (The part of the problem statement, "The length is much greater than its radius," allows to make an approximation: you can ignore the end-caps of the cylinder.)

With that, the left hand side of the equation is simply the electric field's magnitude E, multiplied by the surface area of the Gaussian surface. Here the surface area of the Gaussian surface is the surface area of a cylinder of radius r (ignoring the endcaps). Of course you don't know what E is yet, but that's what you are solving for!

The right hand side of the equation is proportional to the total charge enclosed in the Gaussian surface.

So you need to find what the total charge is within the cylinder, when rR0. The total charge Qenc within the Gaussian surface (where rR0) is the volume charge ρ multiplied by the volume of a cylinder with radius r.
Will it be the same for outside the insulator?
No, it is not quite the same. Use the same approach as the above, but when calculating Qenc, you need to use the radius of the physical cylinder, not the radius of the Gaussian surface.

Thanks Collinsmark

I believe I have it down correctly now,

for inside the cylinder I ended up with E= pr/2ε0

and for outside the cylinder I ended up with E= pR02/r2ε0

Great job!

Yes, for outside the insulator, the electric field can also be found using Gauss's law. The electric field will follow the same formula as inside the cylinder, E= p/(pi*r^2*ε0), where r is the distance from the center of the cylinder. This is because the charge distribution is uniform and the electric field will be radial in all directions. However, the electric field outside the cylinder will be weaker compared to inside the cylinder due to the inverse square relationship with distance.

## 1. What is an electric field in a cylindrical insulator?

The electric field in a cylindrical insulator is a measure of the force per unit charge exerted on a charged particle placed inside the insulator. It is created by the presence of electric charges on the surface or within the insulator.

## 2. How is the electric field in a cylindrical insulator calculated?

The electric field in a cylindrical insulator can be calculated using the formula E = V/d, where E is the electric field, V is the potential difference, and d is the distance between the two points. Alternatively, it can also be calculated using Gauss's Law, which states that the electric field is equal to the charge enclosed divided by the permittivity of the insulator multiplied by 2πr, where r is the radius of the cylindrical insulator.

## 3. What factors affect the strength of the electric field in a cylindrical insulator?

The strength of the electric field in a cylindrical insulator is affected by the magnitude and distribution of electric charges, as well as the permittivity of the insulator. The distance between the two points and the geometry of the insulator also play a role in determining the strength of the electric field.

## 4. How does the electric field in a cylindrical insulator differ from that in a solid insulator?

The electric field in a cylindrical insulator is radial, meaning it points from the center of the insulator towards the outer surface. In contrast, the electric field in a solid insulator is uniform and points in the same direction throughout the insulator. Additionally, the electric field in a cylindrical insulator is affected by the curvature of the surface, while the electric field in a solid insulator is not.

## 5. What are some practical applications of the electric field in cylindrical insulators?

The electric field in cylindrical insulators is used in various electronic devices and systems. It is used in capacitors, which are essential in electronic circuits. It is also used in electrostatic precipitators, which remove particles from gas streams, and in high voltage transmission lines to transport electricity over long distances without significant power loss.

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