two large, parallel, conducting plates are 10.0 cm apart and have charges of equal magnitude and opposite…

two large, parallel, conducting plates are 10.0 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. an electrostatic force of 5.91×10−15 n acts on an electron placed anywhere between the two plates. find the magnitude of the electric field between the plates.

two large, parallel, conducting plates are 10.0 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. an electrostatic force of 5.91×10−15 n acts on an electron placed anywhere between the two plates. find the magnitude of the electric field between the plates.

Answer

Explanation:

Step1: Recall the formula for electric - field force

The force on a charged particle in an electric field is given by $F = qE$, where $F$ is the force, $q$ is the charge of the particle, and $E$ is the electric field. The charge of an electron is $q=- 1.6\times10^{-19}\ C$.

Step2: Rearrange the formula to solve for $E$

We can rewrite the formula $F = qE$ as $E=\frac{F}{q}$.

Step3: Substitute the given values

We are given that $F = 5.91\times10^{-15}\ N$ and $q = 1.6\times10^{-19}\ C$. Substituting these values into the formula $E=\frac{F}{q}$, we get $E=\frac{5.91\times10^{-15}\ N}{1.6\times10^{-19}\ C}$.

Step4: Calculate the value of $E$

$E=\frac{5.91\times10^{-15}}{1.6\times10^{-19}}=\frac{5.91}{1.6}\times10^{-15 + 19}=3.69375\times10^{4}\ N/C$.

Answer:

$3.69\times10^{4}\ N/C$