Question

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.

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:

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