Electric field and charge enclosed

In summary, a cube with sides of 8.10 m is shown in the given diagram. The cube has a non-uniform electric field in the x-direction, with a magnitude of 6.35 N/(Cm^n) where n=1.9. To find the total charge within the cube, the formula q = epsilon * integral (E*dA) is used. However, since the orientation of the cube with respect to the axes is not specified, the sides that contribute to the flux must be determined. Once these sides are identified, the field is evaluated and used to calculate the total flux. The negative and positive contributions of the flux are then taken into account to find the total charge within the cube.
  • #1
2Pac
37
0
Consider a cube of sides a=8.10 m, as shown in the diagram below. Suppose that a non-uniform electric field is present and is given by E=bx^n in the direction of the x-axis, where
b=6.35 N/(Cm^n) and n=1.9. Calculate the magnitude of the total charge within the cube.

so i started by saying
q = epsilon * integral (E*dA)
so q= epsilon * integral (6.35*x^1.9*a^2)dA evaluated from 0 to 8.1m
any help (this is the wrong answer)
 
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  • #2
Since you didn't include the diagram, we don't know how the cube is oriented with respect to the axes. You need to integrate [itex]\epsilon_0 \vec{E} * d\vec{A}[/itex] over the surface of the cube. Assuming the sides of the cube are aligned with the axes, the only sides that have non-zero flux are the two sides that are parallel to the y-z plane; where are they? For example, are they at x_1 = 0 and x_2 = 8.10 m? Evaluate the field on those surfaces (the field will be constant along the surface) to find the total flux. (Note: Signs matter!)
 
  • #3
the cube is along the axes. imagine the origin, then having x-point to your left, y-point to your right and z-point straight up. but I am not sure how to integrate this problem..
 
  • #4
blueskadoo42 said:
the cube is along the axes. imagine the origin, then having x-point to your left, y-point to your right and z-point straight up. but I am not sure how to integrate this problem..
Since the field points in the x-direction, the "integration" should be easy. See my previous suggestions.
 
  • #5
so one will be postive integral of EdA and one will be a negative integral of EdA. wouldn't those just cancel??
 
  • #6
blueskadoo42 said:
so one will be postive integral of EdA and one will be a negative integral of EdA. wouldn't those just cancel??
They would cancel if they had the same magnitude, but do they?
 

1. What is an electric field?

An electric field is a region of space where an electric charge experiences a force. This force can either be attractive or repulsive, depending on the sign of the charge and the direction of the field.

2. How is an electric field calculated?

The electric field at a specific point is calculated by dividing the force experienced by a small test charge at that point by the magnitude of the charge. Mathematically, it is represented as E = F/q, where E is the electric field, F is the force, and q is the charge.

3. What is the relationship between electric field and charge enclosed?

The electric field at a point is directly proportional to the amount of charge enclosed within a closed surface surrounding that point. This means that as the charge enclosed increases, the electric field at that point also increases.

4. What is the concept of Gauss's law in relation to electric field and charge enclosed?

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This allows for easier calculation of the electric field by using symmetry and simplifying the shape of the surface.

5. How does the shape of a charged object affect the electric field and charge enclosed?

The shape of a charged object can affect both the electric field and charge enclosed. For example, a pointed object will have a higher electric field and charge enclosed at its tip compared to a rounded object with the same charge. This is due to the concentration of the charge at the pointed tip.

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