Electric fields through parallelepiped

In summary, the conversation discusses using a Gauss surface to calculate the charge inside a shape, but the speaker is uncertain about a specific situation involving a parallelepiped and infinite parallel plates. They suggest that the fields from the two sources could be combined through superposition to get the correct total fields. Finally, they ask for clarification on what the total flux would be if the field were due to charges outside the parallelepiped.
  • #1
Fluxthroughme
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I have taken my Gauss surface as the front of the shape, with [itex]E_1[/itex] coming through uniformly. I get the right answer for the charge inside the shape, but I'm unsure about b. I imagine a situation that I've drawn could be possible, but I've never seen it before, so I do not know. I'm thinking that the parallelepiped is going through one of the plates involved in an infinite parallel plate set up, with a single infinite plate above them. The two fields would combine as I have labelled, and would, by superposition, give the right total fields at the places they required. If anyone could let me know whether this is fine, or just nonsense, it'd be helpful. Thank you.
 
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  • #2
If the field were due to charges outside the parallelepiped, what would be the total flux?
 

Related to Electric fields through parallelepiped

1. What is an electric field?

An electric field is a physical field that is created by electrically charged particles. It is a vector field, meaning that it has both magnitude and direction, and it exerts a force on other charged particles within its field.

2. How is an electric field created through a parallelepiped?

An electric field through a parallelepiped is created by placing a charged object at one corner of the parallelepiped and measuring the strength and direction of the electric field at various points within the parallelepiped.

3. What factors affect the strength of an electric field through a parallelepiped?

The strength of an electric field through a parallelepiped is affected by the magnitude of the charge, the distance between the charged object and the point being measured, and the dielectric constant of the material within the parallelepiped.

4. How is the direction of an electric field through a parallelepiped determined?

The direction of an electric field through a parallelepiped is determined by the direction of the force that would be exerted on a positive test charge placed at that point within the parallelepiped. The direction of the force is always in the direction of the electric field.

5. What is the unit of measurement for electric field strength?

The unit of measurement for electric field strength is Newtons per Coulomb (N/C) in the SI system and Volts per meter (V/m) in the CGS system.

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