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Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the electric field 4 cm from the plate is:

Respuesta :

skyluke89
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
[tex]E= \frac{\sigma}{2\epsilon_0} [/tex]
where
[tex]\sigma[/tex] is the charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
[tex]E=30 N/C[/tex]

The electric field at 4 cm from the plate is will also be 30 N/C because the distance does not affect the intensity of the electric field.

From Gauss law:

For a surface with uniform charge density:

[tex]E = \dfrac {\sigma}{2\epsilon _0}[/tex]

Where

[tex]\sigma[/tex] - charge density

[tex]\epsilon_0[/tex] -  vacuum permittivity

From the equation, the distance does not affect the intensity of the electric field.  Thus, the electric field 2 cm or 4 cm away from the plate will be equal.

Therefore, the electric field at 4 cm from the plate is will also be 30 N/C.

Learn more about Gauss law,

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