Calculating Electric Flux and Charge: A Problem-Solving Approach

In summary, the conversation is about a closed surface with dimensions a = b = 0.400 m and c = 0.600 m, located a distance "at" from the y-z plane. The electric field is nonuniform and can be found by viewing the attached Microsoft Word Document. In order to receive help on PF, the person must provide their attempted solution and explain how they think the problem can be solved.
  • #1
nosracsan
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1. A closed surface with dimensions a = b = 0.400 m and c = 0.600 m is located as shown in the figure (be aware that the distance “a” in the figure indicates that the rectangular box sits a distance “at” from the y-z plane). The electric field throughout the region is nonuniform and given by:

PLEASE see attached Microsoft Word Document for the rest of the problem... its easier to view in there!
 

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  • #2
Hi nosracsan

In order to receive help on PF, you must

(1) tell us how you think the problem could be solved and
(2) show us your attempted solution.

(forum rules)

The same goes for your thread on point charges.
 
  • #3


I would approach this problem by first understanding the concept of electric flux and total charge. Electric flux is a measure of the electric field passing through a given surface, while total charge is the sum of all the electric charges within that surface.

To find the electric flux passing through the closed surface in this problem, I would use the formula Φ = ∫∫E⃗ · dA⃗, where Φ represents the electric flux, E⃗ is the electric field vector, and dA⃗ is the differential area element. Since the electric field is nonuniform, I would need to break down the surface into smaller areas and calculate the flux for each one, then add them up to get the total flux.

Next, to find the total charge within the closed surface, I would use the formula Q = ε0∫∫∫ρdV, where Q represents the total charge, ε0 is the permittivity of free space, and ρ is the charge density. Again, since the electric field is nonuniform, I would need to calculate the charge density for each small volume element within the surface and then integrate over the entire volume to get the total charge.

To solve this problem, I would need to know the specific values of the electric field and charge density at each point within the surface. This information can be obtained through experiments or simulations. I would also need to know the orientation of the surface and the direction of the electric field, as these factors can affect the calculation of electric flux.

In summary, finding the flux and total charge for a nonuniform electric field requires breaking down the surface into smaller areas or volumes and using the appropriate formulas to calculate the flux and charge density at each point. This problem highlights the importance of accurately measuring and understanding the electric field in a given region, as it directly affects the flux and total charge within a closed surface.
 

FAQ: Calculating Electric Flux and Charge: A Problem-Solving Approach

1. What is flux and how is it related to total charge?

Flux is a measure of the flow of an electric field through a surface. It is directly proportional to the total charge enclosed within the surface.

2. How do you calculate flux?

Flux can be calculated by taking the dot product of the electric field and the surface area vector. This can be represented by the equation: flux = E * A * cos(theta), where E is the electric field, A is the surface area, and theta is the angle between the two vectors.

3. What is the SI unit for flux?

The SI unit for flux is the volt-meter squared (V*m2).

4. What is the relationship between flux and Gauss's law?

Gauss's law states that the total electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space. This relationship allows us to use flux to find the total charge enclosed within a surface.

5. Can flux have a negative value?

Yes, flux can have a negative value if the electric field and surface area vectors are in opposite directions. This indicates that the flow of the electric field is in the opposite direction of the surface area vector.

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