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Electromagnetism & optics

2.
a. A fixed volume charge distribution of constant charge density p_0 is contained within a rectangular box centered at the origin of a Cartesian coordinate system (x, y, z). The box has dimensions w X w X d, where d << w ( i.e. you can model the sheet as infinite in the x and y directions).
i) Find the total charge inside the box
ii) Find the electric field E(z) on the z-axis above, below and inside the box under the assumption that z << w.
iii) On the z-axis, find the difference in the electric potential between the bottom and center of the box i.e., find V(z = 0) - V(z = -d/2), and between the bottom and the top of the box i.e., find V(z = d/2) -V(z = -d/2).

b. A large plane parallel slab of a linear homogeneous dielectric material also of thickness f with a relative permittivity, E_p is placed on the top surface of the charged box of question 2(a).
i) Find the electric field inside the diaelectric
ii) Fine the polarization vector P on the z-axis
iii) Find the bound volume and surface polarization charge densities along the z- axis

c. Consider a charge distribution enclosed in a box with the same geometry as the box in question 2(a) and 2(b), but rather than having a constant charge density, the charge density varies inside the box with z as p(z) = k(z + d/2), where k is a constant.
i) Fine the total charge inside the box.
ii) Find the electric field E(z) on the z-axis inside the box.
iii) On the z-axis, find the difference in the electric potential between the bottom and center of the box, i.e., find V(z = 0) - V(z = -d/2), and between the bottom and the top of the box, i.e., find V(z = d/2) - V(z = -d/2).

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2. (a) A fixed volume charge distribution of constant charge density is contained within a rectangular box centered at the origin of a Cartesian coordinate system. The box has dimensions , where

(i) We wish to find the total charge inside the box.

The total charge is given by

where is the volume of the box.

(ii) We wish to find the electric field on the z-axis above, below, and inside the box.

To compute , we construct a small Gaussian pillbox containing the z-axis with one flat face at , the other flat face at , and the curved face parallel to the z-axis, and thus to the electric field. By symmetry, we have

Thus we see that the electric flux through is zero. The electric flux through is also zero since the electric field is parallel to . Thus by Gauss' law, we have

where A is the area of (and ). Now ...

Solution Summary

We solve various problems involving the electric field, displacement field, and polarization of dielectric materials in the presence of electric charges.

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