Parallelepiped with two electric fields. Find the charge inside.

[bergen89]

Homework Statement

The electric field E1 is constant over the whole of one surface of
parallelepiped and directed out of the surface. On the opposite surface is
field E2 constant over the surface and is directed into the surface. the two
surfaces forming an angle of 30 ° with the horizontal, while both E1 Vector
and E2 vector is horizontal.

Calculate the charge q.

E1: 2,50 x 10^4 Newton Columb
E2: 7.00 x 10^4 Newton Columb

Homework Equations

E1: 2,50 x 10^4 Newton Columb
E2: 7.00 x 10^4 Newton Columb

The Attempt at a Solution

I have added a PDF file With my attempt at a solution. I used Gaus Law to now becouse that's the only thing I can get out of the task. I be super happy if someone could explain how to do this task in a way I hopefully will understand :)

 

[mighty2000]

Hello,

Thank you for sharing your attempt at a solution. Your use of Gauss' Law is a good start, but there are a few things that can be improved upon.

Firstly, it is important to note that the electric field is a vector quantity, meaning it has both magnitude and direction. In this problem, we are given the magnitudes of E1 and E2, but not their directions. It is important to specify the directions of these fields in order to accurately calculate the net electric field and ultimately the charge.

Secondly, the statement mentions that the surfaces of the parallelepiped are at an angle of 30° with the horizontal, but it does not specify the orientation of the parallelepiped itself. This is important because it will affect the calculation of the net electric field.

In order to solve this problem, we can use the following steps:

1. Draw a diagram of the situation, including the direction and orientation of the electric fields and the angle between the surfaces of the parallelepiped.

2. Use the given magnitudes of E1 and E2 to calculate the components of each field in the horizontal direction (E1x and E2x) and in the vertical direction (E1y and E2y). This can be done using trigonometry.

3. Use the components of each field to calculate the net electric field in the horizontal direction (Ex) and in the vertical direction (Ey). This can be done by adding the components of each field together.

4. Use the magnitude and direction of the net electric field to calculate the charge enclosed by a Gaussian surface that includes both surfaces of the parallelepiped. This can be done using Gauss' Law.

I hope this helps guide you in the right direction. If you have any further questions or need more clarification, please don't hesitate to ask.