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Electric field slab, nonuniform charge density

  • Thread starter AKJ1
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1. Homework Statement

A rather large non conducting slab of area A and thickness d has a charge density given by ρ = αx2.
The origin is through the center of the slab. That is, it bisects the slab into two equal volumes of d/2 thickness and with an area A, with -L/2 to the left of x=0 and L/2 to the right of x=0.

A. Gaussian surface (Cylinder) is positioned such that its volume encompasses the charge contained within the slab. Apply Gauss's Law to the cylinder to determine the electric field to the left and to the right of the slab.

B. Consider another cylinder that is located such that its left face at x = 0 and its right face is outside the slab at x>L/2. Apply Gauss's Law to determine the electric field to the right of the slab.

2. Homework Equations

E⋅dA = Qenc / ε0

ρ = Q/V


3. The Attempt at a Solution

A.

dq = ρdV = (αx2)(Adx) [ I integrated this from 0→L/2]

Q = (AαL3 )/ 24

E is constant for this configuration

EA = Q/ε0

E = αL3/ (24ε0)

Now I realize I need to take into account the other half of the slab and the "domain" on x. That is, it will be in the positive x hat direction for x > 0. But this is my question..

B.


When I went to do part B, I don't see how my set-up or my solution will change much from the above. In fact, I got the same magnitude for the electric field.

Did I make a mistake somewhere?
 

Simon Bridge

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Does the electric field due to an assembly of charge depend on the method used to calcukate it?
 
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Does the electric field due to an assembly of charge depend on the method used to calcukate it?
Good point :smile:
 

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