Trying to derive Gauss' law using a cylindrical surface

In summary, the conversation discusses deriving Gauss's law with a straight line of charge and a cylindrical surface. The formula for the electric field of a line of charge is provided, as well as the formula for the surface area of a cylinder. It is mentioned that the product of the electric field and the surface area does not equal the enclosed charge divided by the permittivity. The person asks for clarification on what they are doing wrong. The expert then explains that the problem involves an infinitely long and homogeneous line of charge, and suggests using an arbitrary cylindrical volume and carefully evaluating Gauss's Law to determine the electric field.
  • #1
annamal
381
33
When I try to derive Gauss's law with a straight line of charge with density ##\lambda## through a cylindrical surface of length L and radius R,
$$\vec E = \frac{\lambda*L}{4\pi\epsilon*r^2}$$
$$A = 2\pi*r*L$$
$$\vec E*A = \frac{\lambda *L^2}{2\epsilon*r} \neq \frac{q_{enc}}{\epsilon}$$
What am I doing wrong?
Screen Shot 2022-04-24 at 2.30.39 PM.png
 
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  • #2
annamal said:
$$\vec E = \frac{\lambda*L}{4\pi\epsilon*r^2}$$
Don't confuse the field from a point charge with that of a line of charge. Look it up!
 
  • #3
First of all your "syntax checker" should alarm you that you have equated a vector with a scalar!

I guess you are considering an "infinitely long" homogeneous line-charge along the ##z## axis of a cylindrical coordinate system. Now think about the symmetry of the problem, i.e., in which direction must the electric field point, and then use an arbitrary cylindrical volume, ##V##, with the axis along the line-charge distribution and evaluate carefully Gauss's Law,
$$\int_{\partial V} \mathrm{d}^3 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} Q_V,$$
where ##\partial V## is the boundary of the cylinder (with the surface-normal vectors pointing out of this cylinder), and ##Q_V## is the charge inside the considered cylindrical volume, ##V##.
 

Related to Trying to derive Gauss' law using a cylindrical surface

1. What is Gauss' law and why is it important?

Gauss' law is a fundamental law in electromagnetism that relates the electric field at a point to the charge enclosed within a Gaussian surface surrounding that point. It is important because it allows us to calculate the electric field at a point without having to consider the specific distribution of charges around it.

2. How is a cylindrical surface used to derive Gauss' law?

A cylindrical surface is used to derive Gauss' law by enclosing a certain amount of charge within the surface and then calculating the electric flux through the surface. By manipulating the equations, we can derive Gauss' law which relates the electric field to the enclosed charge.

3. Can Gauss' law be applied to any type of charge distribution?

Yes, Gauss' law can be applied to any type of charge distribution as long as the surface chosen is appropriate for the distribution. For example, a spherical surface is often used for point charges, while a cylindrical surface is used for line charges.

4. What are some real-world applications of Gauss' law?

Gauss' law has many real-world applications, including calculating the electric field around a charged conductor, determining the electric field inside a parallel plate capacitor, and understanding the behavior of lightning and thunderstorms.

5. Are there any limitations to using Gauss' law?

While Gauss' law is a powerful tool for calculating electric fields, it does have some limitations. It can only be used in situations where the electric field is constant and the charge distribution is symmetric. Additionally, it does not take into account the effects of non-electrostatic forces such as magnetic fields.

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