two parallel plates have an area of 0.188 m². how much charge must you place on them to create a 37000 n/c…

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.

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