Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of -8.10 μC/m2 , and sheet B, which is to the right of A, carries a uniform charge of -13.0 μC/m2 . Assume the sheets are large enough to be treated as infinite.Part AFind the magnitude of the net electric field these sheets produce at a point 4.00 cm to the right of sheet A.Part BFind the direction of the net electric field these sheets produce at a point 4.00 cm to the right of sheet A.Part CFind the magnitude of the net electric field these sheets produce at a point 4.00 cm to the left of sheet A.Part DFind the direction of the net electric field these sheets produce at a point 4.00 cm to the left of sheet A.Part EFind the magnitude of the net electric field these sheets produce at a point 4.00 cm to the right of sheet B.

The electric field is a vector magnitude created by the presence of electric charge and is calculated at a point in space. One of the most practical method of calculating the electric field for a system with some symmetry is Gauss Law.

Ф = ∫ E . dA = q_{int} / ε₀

Thr bold indicate vectors, where Ф is the electric flux, E the electric field, A the area, q_{int} charge inside a Gaussian surface and eo the permittivity of the vacuum (ε₀ = 8.85 10⁻¹² C).

For this exercise we have a very large sheet, let's define a Gaussian surface in the form of a cylinder with the base parallel to the surface, in this condition the electric field vector and the normal vector to the bases of the cylinder are parallel, so the product scalar reduces to the algebraic product. In addition, the field is emitted towards both sides of the sheet

E 2A = [tex]\frac{q_{int}}{2 \epsilon_o }[/tex]

If we define a surface charge density

σ = q_{int}/A

E = [tex]\frac{\sigma}{2 \epsilon_o}[/tex]

we can see that the value of the electric field is independent of the distance, so we can reduce the vector sum to an algebraic sum, in the adjoint we can see a diagram with the directions of the electric field for each point.

For the measurements, let's fix a reference system located on sheet A with the positive direction to the right.

a) point of interest x = 4.00cm

this point is between the two plates, from the attached diagram the total bonnet is

E_a = E₁ - E₂

E_a = [tex]- \frac{\sigma_a}{2 \epsilon_o} + \frac{\sigma_b}{2 \epsilon_o}[/tex]

E_a = [tex]\frac{- \sigma_a + \sigma_b}{2\ \epsilon_o}[/tex]

let's calculate

E_a = [tex]\frac{(-8.1 + 13.0) 10^{-6} }{ 2 \ 8.85 \ 10^{-12}}[/tex]

E_a = 2.77 10⁵ N / C

b) the positive sign indicates that the field is directed to the right

c) point of interest x = -4.00 cm

in this case the point is to the left of sheet a, in the diagram we can see the direction of the electric field

E_b = E₁ + E₂

E_b = [tex]\frac{(8.1 +13.0)10^{-6} }{2 \ 8.85 \ 10^{-12}}[/tex]

E_b = 1.19 10⁶ N / C

d) The positive sign indicates that the field is to the right

e) point of interest 4.00 cm to the right of sheet B, the distance is

x = 5.00 + 4.00 = 9.00 cm

In the diagram we see the direction of the electric fields at this point

E_c = - (E₁ + E₂)

E_c = - 1.19 10⁶ N / C

the negative sign indicates that the field is directed to the left

In conclusion, using that the electric field is a vector magnitude and Gauss law, we can find the magnitude and direction of the electric field for the points:

a) E_a = 2.77 10⁵ N / C

b) field is directed to the right

c) E_b = 1.19 10⁶ N / C

d) field is to the right

e) E_c = - 1.19 10⁶ N / C, field is directed to the left

learn more about the electric field here:

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