3. (10 marks) (A Non-uniformly Charged Slab: Planar Symmetry) A slab of insulati

Question

3. (10 marks) (A Non-uniformly Charged Slab: Planar Symmetry) A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x- d andx--d. The y- and z- dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a non-uniform charge density given by: where po is a positive constant. Use Gauss's law to find the electric field due to the slab (magnitude and direction) at all points in space

Explanation / Answer

Because of the geometry of the problem, the electric field will be perpendicular to the slabs.

Also, because of the symmetry of the charge distribution relative to x=0, E(x) = - E(-x).
Specifically this impleis taht E(0) = 0.

Now take a "pill box" with the bottom in the yz-plane and the lid at x < d.
Then the only contribution of electric flux will be through the top of the pillbox, and Gauss law gives:

E(x) * A = A * 1/e0 * Integral{ rho(x') dx' }
= A p_o/(e0d^2) Integral{ x'^2 dx'} = A p_0 (x^3 / 3) /(e0 d^2)

Therefore

E(x) = p_0 x^3 / (3 e0 d^2)

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