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Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q while plate #2 has a charge −2 Q. Find the magnitude of the electric field at a point P in the gap. Assume that the separation between the plates is very small compared to the dimensions of the plates.

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MechEngineer

Answer:

[tex]E = \frac{3Q}{2A\epsilon_0}[/tex]

Explanation:

By Gauss Law for electric field:

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

Where [tex]\sigma[/tex] is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

[tex] E = E_1 + E_2[/tex]

[tex] E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}[/tex]

[tex]E = \frac{3Q}{2A\epsilon_0}[/tex]

ogorwyne

3σ/2Aεο would be the magnitude of the electric field.

This has to be solved using Gauss's law.

What is Gauss law?

This is the law in physics that states that the net flux in a closed surface is the same as the electric charge in the closed surface.

The formula is given as

E = σ/2Aεο

The states are said to be parallel and have different charges

Therefore,

E = σ/2Aεο + 2σ/2Aεο

We collect like terms in the equation above.

Then E =  3σ/2Aεο

3σ/2Aεο would be the magnitude of the electric field.

Read more on Gauss law here:https://brainly.com/question/14773637

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