Question

A very long, uniformly charged cylinder has radius R and linear charge density λ. Find the cylinder's electric field strength outside the cylinder,
r≥R
. Give your answer as a multiple of
λ/
ε
0

Answer 1

The electric field of cylinder at a distance $r$ more than radius of cylinder is equal to $\boxed{E=\dfrac{\lambda}{2\pi\epsilon_0r}}}$.

Further Explanation:

Gauss law states that net electric flux through any closed surface is equal to $\dfrac{1}{\epsilon_0}$ times the net electric charge within that closed surface. The hypothetical closed surface is name as Gaussian surface. The electric flux also defined as the total electric field lines, which passes through a surface area of any closed curve.

Given:

The linear charge density of cylinder is $\lambda$.

Concept:

Consider the length of the cylinder is $L$.

Charge on curved surface of cylinder is $\lambda L$.

For a Gaussian curve in the shape of a concentric cylinder (with a radius more than the radius of charged cylinder) we can write gauss law as:

$\phi=\dfrac{q}{\epsilon_0}$

Substitute $\lambda L$ for $q$ in above equation.

$\phi=\dfrac{\lambda L}{\epsilon_0}$                          …… (I)

Here, $\phi$ is the Electric flux and $\epsilon_0$ is the permittivity of vacuum.

Since the electric field is constant for a given distance $r$ from the axis of the cylinder we can write that:

$\phi=E2\pi rL$                                                            …… (II)

Here, $E$ is the electric field of charge on cylinder, $L$ is the height or length of curved surface and $r$ is the radius of Gaussian cylindrical curve.

On equating the equation (I) and equation (II) and Rearranging it for $E$ :

$\boxed{E=\dfrac{\lambda}{2\pi\epsilon_0r}}}$

Thus, the electric field of cylinder at a distance $r$ more than radius of cylinder is equal to $\boxed{E=\dfrac{\lambda}{2\pi\epsilon_0r}}}$.

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Answer Details:

Grade: College

Subject: Physics

Chapter: Electrostatics

Keywords:

Very long, uniformly charged cylinder, radius R, linear charge density λ, cylinder, electric field strength, gauss law, net electric charge and distance r from the axis of the cylinder.

Answer 2

The cylinder's electric field magnitude, at a distance r from the axis of the cylinder (greater than the cylinder's radius), is equal to $E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

Further explanation

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a charge, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density $\lambda$, this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

$\Phi = \frac{\lambda \cdot L}{\epsilon_0}$

Where $\Phi$ is called the Electric flux. Since the electric field is constant for a given distance r from the axis of the cylinder we can write that:

$\Phi = E \cdot 2\pi r \cdot L$

Joining both our expressions we can get that:

$E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

Learn more

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Keywords

Electrons, protons, electric field, cylinder, electric flux