91Ó°ÊÓ

Plate area of capacitor 1,${\mathrm{A}}_1=1.5{\mathrm{cm}}^2\left(\frac{{10}^{-4}{\mathrm{m}}^2}{1{\mathrm{cm}}^2}\right)=1.5×{10}^{-4}{\mathrm{m}}^2$

The electric field between the plates of capacitor 1,${\mathrm{E}}_1=2000\mathrm{V}/\mathrm{m}$

Plate area of capacitor 2,${\mathrm{A}}_2=0.70{\mathrm{cm}}^2\left(\frac{{10}^{-4}{\mathrm{m}}^2}{1{\mathrm{cm}}^2}\right)=7.00×{10}^{-5}{\mathrm{m}}^2$

The electric field between the plates of capacitor 2,${\mathrm{E}}_2=1500\mathrm{V}/\mathrm{m}$

The charge flowing through any surface can be defined as the surface charge density of the surface for the given area. Thus, using this concept the total charge can be given. Now, from the relation of the divergence of the electric field at a point in space that is equal to the charge density divided by the permittivity of space.

Formulae:

The charge on a body due to its surface charge density, $\mathrm{q}=\frac{\mathrm{σ}}{\mathrm{A}}$ (1)

The electric field at a point in space,$\mathrm{E}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$ (2)

The total charge on the two capacitors can be calculated using equation (1) and equation (2) as follows:

$$\begin{gathered} \mathrm{Q}={\mathrm{q}}_1+{\mathrm{q}}_2 \\ ={\mathrm{σ}}_1{\mathrm{A}}_1+{\mathrm{σ}}_2{\mathrm{A}}_2 \\ ={\mathrm{Î\mu }}_0{\mathrm{E}}_1{\mathrm{A}}_1+{\mathrm{Î\mu }}_0{\mathrm{E}}_2{\mathrm{A}}_2 \\ =(8.85×{10}^{-12}\mathrm{F}/\mathrm{m})(2000\mathrm{V}/\mathrm{m})(1.5×{10}^{-4}{\mathrm{m}}^2) \\ +\left(8.85×{10}^{-12}\mathrm{F}/\mathrm{m}\right)\left(1500\mathrm{V}/\mathrm{m}\right)\left(0.7×{10}^{-4}{\mathrm{m}}^2\right) \\ =2.655×{10}^{-12}\mathrm{C}+0.9293×{10}^{-12}\mathrm{C} \\ =3.58×{10}^{-12}\mathrm{C}\left(\frac{1\mathrm{pC}}{{10}^{-12}\mathrm{C}}\right) \\ =3.6\mathrm{pC} \end{gathered}$$

Hence, the value of the charge is 3.6 pC.