two parallel plates of area 1.74·10^(-4) m² have equal and opposite charges of 6.52·10^(-11) c placed on…

Explanation:

Step1: Recall electric - field formula for parallel plates

The electric field $E$ between two parallel plates is given by $E=\frac{\sigma}{\epsilon_0}$, where $\sigma$ is the surface - charge density and $\epsilon_0$ is the permittivity of free space. The surface - charge density $\sigma=\frac{Q}{A}$, so $E = \frac{Q}{A\epsilon_0}$.

Step2: Substitute the given values

We are given that $Q = 6.52\times10^{-11}\text{ C}$, $A=1.74\times 10^{-4}\text{ m}^2$, and $\epsilon_0 = 8.85\times10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2$. [ \begin{align*} E&=\frac{6.52\times 10^{-11}\text{ C}}{(1.74\times 10^{-4}\text{ m}^2)\times(8.85\times 10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2)}\ &=\frac{6.52\times 10^{-11}}{1.74\times8.85\times 10^{-16}}\text{ N/C}\ &=\frac{6.52\times 10^{-11}}{15.399\times 10^{-16}}\text{ N/C}\ &=\frac{6.52}{15.399}\times10^{5}\text{ N/C}\ &\approx0.4246\times 10^{5}\text{ N/C}\ & = 4.25\times 10^{4}\text{ N/C} \end{align*} ]

Answer:

$4.25\times 10^{4}$