a set of charged plates have an area of 8.22*10^-4 m^2 and separation 2.42*10^-5 m. the plates are charged…
a set of charged plates have an area of 8.22*10^-4 m^2 and separation 2.42*10^-5 m. the plates are charged with 5.24*10^-8 c. what is the potential difference v between the plates? (unit = v)
Answer
Explanation:
Step1: Calculate the capacitance
The formula for the capacitance of a parallel - plate capacitor is $C=\frac{\epsilon_0A}{d}$, where $\epsilon_0 = 8.85\times10^{-12}\ F/m$, $A = 8.22\times10^{-4}\ m^2$, and $d=2.42\times10^{-5}\ m$. $C=\frac{8.85\times 10^{-12}\times8.22\times10^{-4}}{2.42\times10^{-5}}$ $C=\frac{8.85\times8.22\times10^{-12 - 4}}{2.42\times10^{-5}}$ $C=\frac{72.747\times10^{-16}}{2.42\times10^{-5}}$ $C = 30.06\times10^{-11}\ F=3.006\times 10^{-10}\ F$
Step2: Calculate the potential difference
The relationship between charge $Q$, capacitance $C$, and potential difference $V$ is $Q = CV$. So, $V=\frac{Q}{C}$. Given $Q = 5.24\times10^{-8}\ C$ and $C = 3.006\times10^{-10}\ F$. $V=\frac{5.24\times10^{-8}}{3.006\times10^{-10}}$ $V=\frac{5.24}{3.006}\times10^{-8 + 10}$ $V\approx174\ V$
Answer:
$174$