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Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.50×105 V/m . When the space is filled with dielectric, the electric field is E= 2.40×105 V/m .

What is the charge density on each surface of the dielectric?

Respuesta :

Explanation:

The given data is as follows.

     Electric field between plates without dielectric, [tex]E_{1} = 3.50 \times 10^{5} V/m[/tex]

   Electric field between the plates with dielectric, [tex]E_{2} = 2.40 \times 10^{5} V/m[/tex].

  Permittivity of free space, [tex]\epsilon_{o}[/tex] = [tex]8.85 \times 10^{-12} C^{2}/Nm^{2}[/tex]

Now, we will determine the charge density as follows.

            [tex]\sigma_{i} = \epsilon_{o}(E_{1} - E_{2})[/tex]

                 = [tex]8.85 \times 10^{-12} \times (3.50 \times 10^{5} - 2.40 \times 10^{5})[/tex]

                 = [tex]9.735 \times 10^{-7} C/m^{2}[/tex]

Thus, we can conclude that the charge density on each surface of the dielectric is [tex]9.735 \times 10^{-7} C/m^{2}[/tex].

onyebuchinnaji

The charge density on each surface of the dielectric is [tex]9.735\times 10^{-7} \ C/m^2[/tex]

The given parameters;

  • electric field between the plates when they are empty, E₁ = 3.5 x 10⁵ V/m
  • electric between the plates with dielectric material, E₂ = 2.4 x 10⁵ V/m

The charge density on each surface of the dielectric is calculated as follows;

[tex]\sigma = \varepsilon_o (E_1 - E_2)\\\\\sigma = 8.85\times 10^{-12} (3.5 \times 10^{5} \ - \ 2.4 \times 10^5)\\\\\sigma = 9.735 \times 10^{-7} \ C/m^2[/tex]

Thus, the charge density on each surface of the dielectric is [tex]9.735\times 10^{-7} \ C/m^2[/tex]

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