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A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface charge density of -2.6×10-4 C/m2 , while there is no excess charge elsewhere in the cloud. What is the magnitude of the electric field, in newtons per coulomb, a few meters below the cloud, under the clouds approximate center, due to the charge distribution?

Respuesta :

skyluke89

Magnitude of the electric field is [tex]1.47\cdot 10^7 N/C[/tex]

Explanation:

In this problem, the length of the cloud (horizontally) is a few kilometers, while here we want to evaluate the electric field it produces just a few meters below the cloud.

Therefore, we can approximate the cloud as being an infinite sheet with charge on it.

The electric field produced by an infinite sheet of charge is perpendicular to the sheet and it is given by

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

where

[tex]\sigma[/tex] is the surface charge density

[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity

For the cloud, we have

[tex]\sigma=-2.6\cdot 10^{-4} C/m^2[/tex]

Therefore, the electric field produced by the cloud is

[tex]E=\frac{-2.6\cdot 10^{-4}}{2(8.85\cdot 10^{-12})}=-1.47\cdot 10^7 N/C[/tex]

Where the negative sign means the field points towards the cloud.

Therefore, the magnitude of the field is

[tex]1.47\cdot 10^7 N/C[/tex]

Learn more about electric fields:

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Lanuel

The magnitude of the electric field in Newtons per coulomb is equal to [tex]1.47 \times 10^7\; N/C[/tex]

Given the following data:

  • Surface charge density = [tex]-2.6\times 10^{-4}\; C/m^2[/tex]

Scientific data:

  • Vacuum permittivity = [tex]8.85 \times 10^{-12}\;F/m[/tex]

To calculate the magnitude of the electric field in Newtons per coulomb:

Note: The magnitude of the electric field that is produced by an infinite cloud of charge is perpendicular to the cloud.

The formula for electric field.

Mathematically, the magnitude of an electric field is given by this formula:

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

Where:

  • [tex]\sigma[/tex] is the surface charge density.
  • [tex]\epsilon_o[/tex] is the vacuum permittivity.

Substituting the given parameters into the formula, we have;

[tex]E=\frac{-2.6\times 10^{-4}}{2\; \times \;8.85 \times 10^{-12}}\\\\E=\frac{-2.6\times 10^{-4}}{1.77 \times 10^{-11}}\\\\E = 1.47 \times 10^7\; N/C[/tex]

Note: The negative sign indicates that the electric field points to the cloud.

Read more on surface charge density here: https://brainly.com/question/8966223

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