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

Answer :

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:

[tex]E=\frac{Q}{A\cdot\epsilon_0}=\frac{31.96\cdot10^{-6}}{13.08\cdot10^{-4}\cdot8.85\cdot10^{-12}}=0.2761\cdot10^{10}\text{ N/C}[/tex]

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:

[tex]\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}[/tex]

Therefore the force is 0.442 nanoNewtons.

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