Respuesta :
Answer:
Surface charge density, [tex]\sigma=1.77\times 10^{-8}\ C/m^2[/tex]
Explanation:
Given that,
Separation between two parallel conducting plates, d = 1 mm = 0.001 m
Potential difference between plates, V = 2V
Let [tex]\sigma[/tex] is the magnitude of the surface charge density on each plate. Its formula is given by Gauss's law :
[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]
[tex]\sigma=E\times \epsilon_o[/tex]
Since, [tex]E=\dfrac{\Delta V}{d}[/tex]
[tex]\sigma=\dfrac{\Delta V}{d}\times \epsilon_o[/tex]
[tex]\sigma=\dfrac{2}{0.001}\times 8.85\times 10^{-12}[/tex]
[tex]\sigma=1.77\times 10^{-8}\ C/m^2[/tex]
So, the magnitude of the surface charge density on each plate is [tex]1.77\times 10^{-8}\ C/m^2[/tex]. Hence, this is the required solution.
Surface charge density is the product of electric field at a point in space to the permittivity.
The magnitude of the surface charge density on each plate is [tex]1.77\times10^{-8}\rm C/m^2[/tex].
What is surface charge density?
Surface charge density is the product of electric field at a point in space to the permittivity.
It can be given as,
[tex]\sigma =E\varepsilon_0[/tex]
Here, (E) is the electric field at a point in space and ([tex]\varepsilon_0[/tex]) is the permittivity.
As the electric field is the ratio of potential difference to the distance between two point.
Thus the above formula can be given as,
[tex]\sigma =\dfrac{\Delta V}{d}\varepsilon_0[/tex]
Given information-
Two parallel conducting plates are separated by 1 mm.
The charge carried by plates is equal but opposite surface charge densities.
The potential difference between the plates is 2 V.
Put the values in the above formula as,
[tex]\sigma =\dfrac{2}{0.0001} \times8.85\times10^{-12}\\\sigma=1.77\times10^{-8}\rm C/m^2[/tex]
Hence the magnitude of the surface charge density on each plate is [tex]1.77\times10^{-8}\rm C/m^2[/tex].
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