two parallel plates are charged with 3.08·10^(-9) c of charge. what must the area of the plates be to create…

Explanation:

Step1: Recall the formula for electric - field between parallel plates

The electric - field between two parallel plates is given by $E=\frac{\sigma}{\epsilon_0}$, where $\sigma=\frac{Q}{A}$ (surface charge density, $Q$ is the charge on the plate, and $A$ is the area of the plate). So, $E = \frac{Q}{A\epsilon_0}$.

Step2: Rearrange the formula to solve for the area $A$

We can rewrite the formula $E=\frac{Q}{A\epsilon_0}$ as $A=\frac{Q}{E\epsilon_0}$.

Step3: Substitute the given values

We are given that $Q = 3.08\times10^{-9}\ C$, $E = 9220\ N/C$, and $\epsilon_0=8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}$. [ \begin{align*} A&=\frac{3.08\times 10^{-9}\ C}{9220\ N/C\times8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}}\ &=\frac{3.08\times 10^{-9}}{9220\times8.85\times 10^{-12}}m^{2}\ &=\frac{3.08\times 10^{-9}}{8.15\times10^{-8}}m^{2}\ & = 0.0378\ m^{2} \end{align*} ]

Answer:

$0.0378$