# Gauss' Law Help don't even know where to start

• miniMitts27
In summary, the conversation discusses finding the outward flux through each end of a cylinder placed in an electric field. The electric field has a magnitude of 250 N/C and is positive for x > 0 and negative for x < 0. The cylinder is 20 cm long and has a radius of 3 cm, with its center at the origin and its axis along the x-axis. One end of the cylinder is at x = +10 cm and the other is at x = -10 cm. The method of finding the flux is by finding the area of each circular end and multiplying it by the electric field within that area. The person in the conversation initially tried integrating the surface area of the entire cylinder, but realized they only needed
miniMitts27
An electric field is E = 250 N/C i for x > 0 and E = -250 N/C i for x < 0. A cylinder of length 20 cm and radius 3 cm has its center at the origin and its axis along the x-axis such that one end is at x = +10 cm and the other is at x = -10 cm.

What is the outward flux through each end?
a) +x end
b) -x end

I tried integrating the surface area of the cylinder, then multiplying by the electric field within that area but it's not correct... so a starting point or walk-through would be greatly appreciated.

miniMitts27 said:
I tried integrating the surface area of the cylinder,
What do you mean? Did you find the area of each circular end surface?
then multiplying by the electric field within that area but it's not correct...
That should work, assuming you have the correct signs.

Show what you did.

I figured it out actually. They were just looking for the flux through the very end, so I only needed to find the area of the end of the cylinder, then multiply that by the electric field. I was integrating the surface area of the entire cylinder and then multiplying the charge by that. Obviously a very different number.

## What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism, named after German mathematician and physicist Carl Friedrich Gauss. It states that the electric flux through a closed surface is proportional to the enclosed electric charge.

## Why is Gauss' Law important?

Gauss' Law is important because it helps us understand the behavior of electric fields and how they are affected by electric charges. It also allows us to calculate the electric field at any point in space, which is crucial in many practical applications such as designing electronic devices.

## How do you apply Gauss' Law to a specific situation?

To apply Gauss' Law, you need to first identify the closed surface that encloses the electric charge. Then, you can calculate the electric flux through that surface by taking the dot product of the electric field and the surface area. Finally, you can use the equation Q = ε0E*A to solve for the electric field at any point within the closed surface.

## What is the difference between Gauss' Law and Coulomb's Law?

Gauss' Law and Coulomb's Law both deal with the behavior of electric charges, but they approach it in different ways. Coulomb's Law gives the magnitude and direction of the electric force between two point charges, while Gauss' Law relates the electric field to the enclosed charge and the shape of the closed surface.

## Can Gauss' Law be applied to situations with non-uniform electric fields?

Yes, Gauss' Law can be applied to situations with non-uniform electric fields. However, the closed surface used in the calculation must be carefully chosen to ensure that the electric field is constant over each small area element on the surface. This allows us to use the equation Q = ε0E*A to calculate the electric field at any point within the closed surface.

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