Question

two large, parallel, conducting plates are 50.000 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. an electrostatic force of 5.6059 × 10^(-15) n acts on an electron placed anywhere between the two plates. find the magnitude of the potential difference between the plates (neglect fringing).

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 $q=- 1.6\times10^{-19}\ C$.
$E=\frac{F}{q}$

Step2: Calculate the electric field

We know that $F = 5.6059\times10^{-15}\ N$ and $q = 1.6\times10^{-19}\ C$.
$E=\frac{5.6059\times10^{-15}}{1.6\times10^{-19}}=\frac{5.6059}{1.6}\times10^{4}\ V/m=3.5037\times10^{4}\ V/m$

Step3: Recall the formula for potential - difference

The potential - difference $V$ between two parallel plates separated by a distance $d$ in a uniform electric field $E$ is given by $V = Ed$. The distance between the plates $d = 50.00\ cm=0.5000\ m$.
$V=E\times d$

Step4: Calculate the potential - difference

Substitute $E = 3.5037\times10^{4}\ V/m$ and $d = 0.5000\ m$ into the formula.
$V=(3.5037\times10^{4})\times0.5000 = 1.75185\times10^{4}\ V$

Answer:

$1.75185\times10^{4}\ V$