Electric field and charge enclosed

AI Thread Summary
The discussion revolves around calculating the total charge within a cube subjected to a non-uniform electric field defined by E=bx^n. The initial approach using the formula q = epsilon * integral (E*dA) is critiqued for not considering the correct orientation of the cube and the need to evaluate the electric field on the surfaces parallel to the y-z plane. It is clarified that only these surfaces contribute to the total flux, and the integration should account for the varying electric field across the cube's surfaces. The importance of evaluating the field at both ends of the cube is emphasized, as they will not necessarily cancel out due to differing magnitudes. Proper integration and consideration of the electric field's direction are essential for an accurate calculation of the enclosed charge.
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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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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 \epsilon_0 \vec{E} * d\vec{A} 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!)
 
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..
 
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.
 
so one will be postive integral of EdA and one will be a negative integral of EdA. wouldn't those just cancel??
 
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?
 

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