A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $$3 \times 10^4$$ V/m, the charge density of the positive plate will be close to:

We have a parallel-plate capacitor whose space between the plates is completely filled with a dielectric material of dielectric constant $$\kappa = 2.2$$.

For such a situation the free (i.e. plate) surface charge density $$\sigma$$ is related to the electric field $$E$$ inside the dielectric through the electric displacement vector $$\vec D$$.

By definition, for an isotropic dielectric

$$\vec D \;=\; \varepsilon_0 \kappa \,\vec E$$

and the normal component of $$\vec D$$ at the surface of a conductor equals the free charge density on that surface, so

$$\sigma \;=\; D_n \;=\; \varepsilon_0 \kappa E.$$

Now we substitute the numerical values.

The permittivity of free space is

$$\varepsilon_0 = 8.854 \times 10^{-12}\;{\rm F\,m^{-1}}.$$

The given electric field is

$$E = 3 \times 10^{4}\;{\rm V\,m^{-1}}.$$

Hence

$$\sigma \;=\; \varepsilon_0 \kappa E$$

$$\;=\; (8.854 \times 10^{-12}) \times (2.2) \times (3 \times 10^{4}).$$

Carrying out the multiplication step by step, we first multiply $$\varepsilon_0$$ with $$\kappa$$:

$$8.854 \times 10^{-12} \times 2.2 \;=\; 19.4788 \times 10^{-12}$$

$$\;=\; 1.94788 \times 10^{-11}.$$

Now we multiply this by the electric field:

$$1.94788 \times 10^{-11} \times 3 \times 10^{4}$$

$$\;=\; 5.84364 \times 10^{-7} \;{\rm C\,m^{-2}}.$$

This value is practically $$5.8 \times 10^{-7}\;{\rm C\,m^{-2}}$$, which is closest to $$6 \times 10^{-7}\;{\rm C\,m^{-2}}$$ among the options provided.

Hence, the correct answer is Option A.