Question

two large, parallel, conducting plates are 10.0 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. the plates are oriented so that an electrostatic force of 5.64×10−15 n acts to the right on an electron placed anywhere between the two plates. (neglect fringing.) find the magnitude of the electric field between the plates. enter a positive number if the field points to the left and a negative number if the field points to the right.

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 strength. 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 values of $F$ and $q$

We are given that $F = 5.64\times10^{-15}\ N$ and $q=-1.6\times10^{-19}\ C$. Substituting these values into the formula $E=\frac{F}{q}$, we get $E=\frac{5.64\times10^{-15}}{- 1.6\times10^{-19}}$.

Step4: Calculate the value of $E$

$E=\frac{5.64\times10^{-15}}{-1.6\times10^{-19}}=\frac{5.64}{-1.6}\times10^{-15 + 19}=-3.525\times10^{4}\ N/C$. If the field points to the left, we enter a positive number. So the magnitude of the electric field is $E = 3.525\times10^{4}\ N/C$.

Answer:

$3.525\times10^{4}$