From Gauss's law, the electric field is expressed as:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{{\mathbf{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{d}}}{\mathbf{4}{\mathbf{\pi \epsilon}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}}$

The electric field at point P due to the circular disk is given by:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{\sigma}\mathbf{z}}{\mathbf{2}{\mathbf{\epsilon}}_{\mathbf{0}}}{\mathbf{(}}\frac{\mathbf{1}}{\mathbf{z}}{\mathbf{+}}\frac{\mathbf{1}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}}{\mathbf{)}}}$, where σ is the charge density, z is the distance from a point to the disk, and R is the radius of the disk.

The electric field due to a conducting surface is:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{\sigma}}{{\mathbf{\epsilon}}_{\mathbf{0}}}}$

The surface charge density, σ is the ratio of the total charge to the surface area.

That is:

$\overline{){\mathbf{\sigma}}{\mathbf{=}}\frac{{\mathbf{Q}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}}{\mathbf{A}}}$

**A)**

The aluminum foil is 3mm from the plastic disk.

Therefore, the charge will occupy the surface of the aluminum foil due to the charge on the plastic disk.

A large, thin plastic disk with radius R = 1.5 meter carries a uniformly distributed charge of Q = -3x10^{-5} C. A circular piece of aluminum foil is placed d = 3 mm from the disk, parallel to the disk. The foil has a radius of r = 2 cm and a thickness t = 1 millimeter.

A) Show the charge distribution on the close-up foil

B) Calculate the magnitude and direction of the electric field at the center of the foil, inside the foil.

C) Calculate the magnitude q of the charge on the left circular face of the foil.

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