Question

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?

Answer 1

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

`\vec{E}=(Q)/(A\epsilon_0)`

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

`E=(20.93*10^(-6))/((32.93*10^(-4))*(8.85*10^(-12)))=718181521.8(V)/(m)`

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

`F=q.E`

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

Finally, our force can be written as:

`F=1.6*10^(-19)*718181521.8=1.149*10^(-19)=0.00001149microN`

Our final answer is 0.00001149 micro Newtons