91影视

The given data can be listed below,

• The electric field of air is,$E=3脳{10}^6\mathrm{N}/\mathrm{C}$
• The radius of capacitor is,$R=47cm$
• The gap between two disk of capacitor is,$d=1\mathrm{mm}$

When the voltage is supplied across the capacitor's metallic plates, a charge is stored electrostatically.

A capacitor is made up of two metallic plates separated by a dielectric substance. The charge is stored in the dielectric substance, which is non-conductive.

The electric field on the capacitor is given by,

$E=\frac{q}{A{蔚}_0}$

Here, q is the charge on the disk, A is the area of disk and ${蔚}_0$is the permittivity of free space.

Substitute all the values in the above equation, the charge on the disk is given by,

$q=EA{蔚}_0$

role="math" localid="1656930383038" $$\begin{gathered} =\left(3脳{10}^6\mathrm{N}/\mathrm{C}\right)蟺{\left(0.47\mathrm{m}\right)}^2\left(8.85脳{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right) \\ =1.84脳{10}^{-5}\mathrm{C} \end{gathered}$$

Thus, the maximum charge that can be placed on disk is role="math" localid="1656930441071" $1.84脳{10}^{-5}\mathrm{C}$.