An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
.7 μC/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.8 cm and b = 4.8 cm. The conducting slab has a net charge per unit area of σ2 = 88 μC/m2. What is Ex(P), the value of the x-component of the electric field at point P, located a distance 6.8 cm from the infinite sheet of charge?
We use the Gauss Law and the superposition law in order to solve this problem.
<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:
<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:
Then:
This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.
If we compare the energy of visible light to the energy of X-rays, we find that X-rays have a much higher frequency. Usually, electromagnetic radiation with higher frequency (energy) have a higher degree of penetration than those with low frequency.