two parallel plates are charged with 4.98·10^(-7) c of charge. what must the area of the plates be to create…

Explanation:

Step1: Recall electric - field formula for parallel plates

The electric - field formula for parallel plates is $E=\frac{\sigma}{\epsilon_0}$, where $\sigma=\frac{Q}{A}$ (surface charge density), $Q$ is the charge on the plates, $A$ is the area of the plates, and $\epsilon_0$ is the permittivity of free space. Substituting $\sigma=\frac{Q}{A}$ into $E = \frac{\sigma}{\epsilon_0}$, we get $E=\frac{Q}{A\epsilon_0}$.

Step2: Solve for the area $A$

Rearrange the formula $E=\frac{Q}{A\epsilon_0}$ to solve for $A$. We have $A=\frac{Q}{E\epsilon_0}$. Given $Q = 4.98\times10^{-7}\ C$, $E = 8720\ N/C$, and $\epsilon_0=8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}$. Substitute the values into the formula: [ \begin{align*} A&=\frac{4.98\times 10^{-7}\ C}{8720\ N/C\times8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}}\ &=\frac{4.98\times 10^{-7}}{8720\times8.85\times 10^{-12}}\ m^{2}\ &=\frac{4.98\times 10^{-7}}{7.7172\times 10^{-8}}\ m^{2}\ &\approx6.45\ m^{2} \end{align*} ]

Answer:

$6.45$