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=13.08 cm2, and a charge of 31.96 microC, what force, in nanoNewtons, would an electron experience due to this electric field?

Answer 1

First, let's convert the area to m² and the charge to Coulomb:

13.08 cm² = 13.08 * 10^-4 m²

31.96 uC = 31.96 * 10^-6 C

The constant "eo" is the vacuum permittivity, equal to 8.85 * 10^-12.

So, calculating the electric field, we have:

`E=(Q)/(A\cdot\epsilon_0)=(31.96\cdot10^(-6))/(13.08\cdot10^(-4)\cdot8.85\cdot10^(-12))=0.2761\cdot10^(10)\text{ N/C}`

Now, to find the force acting on an electron, let's use the formula below, knowing that the charge of an electron is 1.6 * 10^-19 C:

`\begin{gathered} F=qE\\ \\ F=1.6\cdot10^(-19)\cdot0.2761\cdot10^(10)\\ \\ F=0.44176\cdot10^(-9)\text{ N}\\ \\ F=0.442\text{ nN} \end{gathered}`

Therefore the force is 0.442 nanoNewtons.