A parallel plate capacitor with plates of area 1 m$$^2$$ each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is: (Take $$\epsilon_0 = 8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2}$$)

We have two large, parallel conducting plates of a capacitor. Because the plates are very close and very wide, the electric field between them is practically uniform. For such a configuration, electrostatics gives the direct relation between the electric field $$E$$ and the surface charge density $$\sigma$$ on either plate:

$$E \;=\; \dfrac{\sigma}{\epsilon_0}.$$

Here $$\epsilon_0$$ is the permittivity of free space. We are given

$$E = 100 \; \text{N/C}, \qquad \epsilon_0 = 8.85 \times 10^{-12} \; \dfrac{\text{C}^2}{\text{N}\cdot\text{m}^2}.$$

Re-arranging the formula to solve for $$\sigma$$, we write

$$\sigma = \epsilon_0 \, E.$$

Substituting the numerical values,

$$\sigma = \left( 8.85 \times 10^{-12} \; \dfrac{\text{C}^2}{\text{N}\cdot\text{m}^2} \right) \!\times\! \left( 100 \; \dfrac{\text{N}}{\text{C}} \right).$$

Now we multiply the powers of ten and the coefficients step by step:

$$8.85 \times 10^{-12} \times 100 \;=\; 8.85 \times 10^{-12} \times 10^{2} \;=\; 8.85 \times 10^{-10}.$$

Hence,

$$\sigma = 8.85 \times 10^{-10} \; \dfrac{\text{C}}{\text{m}^2}.$$

The plates each have an area

$$A = 1 \; \text{m}^2.$$

The total charge $$Q$$ residing on one plate is the surface charge density times the area:

$$Q = \sigma \, A.$$

Substituting $$\sigma$$ and $$A$$,

$$Q = \left( 8.85 \times 10^{-10} \; \dfrac{\text{C}}{\text{m}^2} \right) \!\times\! \left( 1 \; \text{m}^2 \right) = 8.85 \times 10^{-10} \; \text{C}.$$

This value represents the magnitude of charge on each plate of the capacitor. The separation between the plates (0.1 m) is immaterial in this calculation because the relation $$E = \sigma / \epsilon_0$$ already embodies the correct dependence for an ideal parallel-plate capacitor.

Hence, the correct answer is Option A.