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The electric field from a sheet of charge is perpendicular to the sheet and has a constant magnitude of Q/(Aeo), where A is the area of the sheet and Q is the charge on the sheet. If the sheet has an area, A=32.93 cm2, and a charge of 20.93 microC, what force, in nanoNewtons, would an electron experience due to this electric field?

Respuesta :

The exercise tells us that the electric field is given by the following equation:

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

And it also gives us, A and Q. Thus, our electric field inside the capacitor is:

[tex]E=\frac{20.93*10^{-6}}{(32.93*10^{-4})*(8.85*10^{-12})}=718181521.8\frac{V}{m}[/tex]

As we know, the electric force can be written as:

[tex]F=q.E[/tex]

The charge of an electron is a constant, which is q=1.6*10^(-19) C.

Finally, our force can be written as:

[tex]F=1.6*10^{-19}*718181521.8=1.149*10^{-19}=0.00001149microN[/tex]

Our final answer is 0.00001149 micro Newtons

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