a set of charged plates is separated by 2.22*10^-4 m. when 5.24*10^-9 c of charge is placed on the plates…
a set of charged plates is separated by 2.22*10^-4 m. when 5.24*10^-9 c of charge is placed on the plates, it creates a potential difference of 240 v. what is the area of the plates? (the answer is ____*10^-4 m^2. just fill in the number, not the power.)
Answer
Explanation:
Step1: Calculate the capacitance
The formula for capacitance is $C=\frac{Q}{V}$, where $Q$ is the charge and $V$ is the potential - difference. Given $Q = 5.24\times10^{-9}\ C$ and $V=240\ V$. $C=\frac{5.24\times10^{-9}}{240}\ F$
Step2: Use the capacitance formula for parallel - plate capacitor
The capacitance formula for a parallel - plate capacitor is $C=\frac{\epsilon_0A}{d}$, where $\epsilon_0 = 8.85\times10^{-12}\ F/m$, $A$ is the area of the plates, and $d$ is the distance between the plates. Given $d = 2.22\times10^{-4}\ m$. We can re - arrange the formula for $A$: $A=\frac{C\times d}{\epsilon_0}$ Substitute $C=\frac{5.24\times10^{-9}}{240}$ into the formula for $A$: $A=\frac{\frac{5.24\times10^{-9}}{240}\times2.22\times10^{-4}}{8.85\times10^{-12}}$ $A=\frac{5.24\times10^{-9}\times2.22\times10^{-4}}{240\times8.85\times10^{-12}}$ $A=\frac{5.24\times2.22\times10^{-13}}{240\times8.85\times10^{-12}}$ $A=\frac{5.24\times2.22}{240\times8.85}\times10^{-13 + 12}$ $A=\frac{11.6328}{2124}\times10^{-1}$ $A = 0.005477\times10^{-1}$ $A= 5.48\times10^{-4}\ m^2$
Answer:
5.48