Question

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 and x = -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 uniform positive charge density p.
a. Explain why the electric field due to the slab is zero at the center of the slab (x=0).

b. Using Gauss's law, find the magnitude of the electric field due to the slab at the points 0<= x <= d. Express your answer in terms of the variables p, x, d, and e_0.

d. Using Gauss's law, find the magnitude of the electric field due to the slab at the points x>= d. Express your answer in terms of the variables p, x, d, and e_0.

e. What is the direction of the electric field due to the slab at the pints x >= d.

Answer 1

Answer:

a) In the center of the slab there are no charges inside, so by Gauss's law the electric field is zero,

b)  E = ρ x / 2ε₀   , c)   E = σ / 2ε₀,

d) the direction of the electric field as the charge is positive is leaving the plate  

Explanation:

a) In the center of the slab there are no charges inside, so by Gauss's law the electric field is zero,

Another way of analyzing it is that the charge on one side of the crockery creates an outgoing electric field in the center, the charge on the other side of the crockery creates a field of equal magnitude, but in the opposite direction, so the resulting field is zero. .

b) Let's use Gauss's law to calculate the electric field, let's use as cylinder a Gaussian surface with the base parallel to the faience, so the scalar product is reduced to the algebraic product

         Ф = ∫ E. dA = qint /ε₀

 

The slab area is A

let's use the concept of charge density

         ρ = qint / V

the volume of the slab is the area times the thickness

         V = A x

       qint = ρ A x

as the two sides of the slab create an electric field the flow is

          Φ = E 2A

         

we substitute

       2E A = ρ A x /ε₀  

          E = ρ x / 2ε₀  

where x goes from zero to the thickness of the plate a x = d

c) in the case of x> d

for this case

     all the charge is inside the gaussian surface .  We look for the relationship between volumetric density and surface density

      σ = Q / A

multiply by the thickness d

       σ = q d / Ad = Q d / V

        ρ = Q / V

        σ = ρ d

we see that the product of the density voluntarily by plate thickness is the surface charge density

        E = σ / 2ε₀  

d) to the direction of the electric field as the charge is positive is leaving the plate