two parallel plates have an area of 0.188 m²....
two parallel plates have an area of 0.188 m². how much charge must you place on them to create a 37000 n/c field between the plates? ?·10^? c hint: ε₀ = 8.85·10⁻¹² c²/n·m² enter the coefficient in the green box and the exponent in the yellow one.
Answer
# Explanation:
## Step1: Recall electric - field formula for parallel plates
The electric - field between parallel plates is given by $E=\frac{\sigma}{\epsilon_0}$, where $\sigma=\frac{Q}{A}$ (surface charge density, $Q$ is the charge, and $A$ is the area of the plates). So $E = \frac{Q}{A\epsilon_0}$.
## Step2: Solve for the charge $Q$
Rearranging the formula $E=\frac{Q}{A\epsilon_0}$ for $Q$, we get $Q = E\times A\times\epsilon_0$.
Given $E = 37000\ N/C$, $A=0.188\ m^2$, and $\epsilon_0 = 8.85\times 10^{-12}\ C^2/N\cdot m^2$.
Substitute the values:
\[
\begin{align*}
Q&=(37000\ N/C)\times(0.188\ m^2)\times(8.85\times 10^{-12}\ C^2/N\cdot m^2)\\
&=37000\times0.188\times8.85\times 10^{-12}\ C\\
&=(37000\times0.188\times8.85)\times 10^{-12}\ C\\
&=(37000\times1.6638)\times 10^{-12}\ C\\
& = 61560.6\times 10^{-12}\ C\\
&=6.15606\times 10^{-8}\ C
\end{align*}
\]
# Answer:
6.16, - 8