Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i. e., the slab is uncharged). Required:
Find the magnitude and direction of the electric field.

N/C

N/C

Explanation:

The charge per unit area of the two non-conducting slabs are given by:

The charge density on the metal

ε0 = 8.854 x 10-12 C2/N m2

Note that the electric field inside the conductor is zero because it is an equipotential surface.

The diagram attached to this solution typifies the description given in the question:

The electric field in the region C can be calculated by:

The electric field in the region A can be calculated by:

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

root word, prefixes, and suffixes