91Ó°ÊÓ

The given data is listed below as-

• The given electric field’s magnitude is,$E=3×{10}^{-6}\mathrm{N}/\mathrm{C}$
• The radius of the capacitor is,$R=50\mathrm{cm}=50\mathrm{cm}×\frac{1\mathrm{m}}{100\mathrm{cm}}=0.5\mathrm{m}$
• The gap between the capacitor i$s=1\mathrm{mm}=1\mathrm{mm}×\frac{1\mathrm{m}}{1000\mathrm{mm}}={10}^{-3}\mathrm{m}$

The electric field is described as a region that helps an electrically charged particle to exert force on another particle.

The equation of the charge for the disks is expressed as,

$E=\frac{Q/A}{{\mathrm{Î\mu }}_0}$…(1)

Here, $Q$is the maximum charge placed on the disks, $A$is the area of the disks ($A={\mathrm{Ï€¸é}}^2$) and ${\mathrm{Î\mu }}_0$is the electric field constant.

The equation of the strength for the fringe field is expressed as,

$E_1=\frac{Q}{2A{\mathrm{Î\mu }}_0}·\frac{S}{R}$…(2)

Here, $E_1$is the strength of the fringe and $S$ is the gap between the capacitors

For$E=3×{10}^{-8}N/C$,$R=0.5m$and ${\mathrm{Î\mu }}_0=8.85×{10}^{-12}{C}^2/{\mathrm{Nm}}^2$in equation (1).

$$\begin{gathered} 3×{10}^{-6}\mathrm{N}/\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{π}{\left(0.5\mathrm{m}\right)}^2×8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ \mathrm{Q}=2.65×{10}^{-15}\mathrm{C}/{\mathrm{m}}^2×\mathrm{π}{\left(0.5\mathrm{m}\right)}^2 \\ =2.65×{10}^{-15}\mathrm{C}/{\mathrm{m}}^2×0.785{\mathrm{m}}^2 \\ =2.09×{10}^{-5}\mathrm{C} \\ ≈2.1×{10}^{-5}\mathrm{C} \end{gathered}$$

For $R=0.5m$,$Q=2.09×{10}^{-5}C$ , $S={10}^{-3}\mathrm{m}$and ${\mathrm{Î\mu }}_0=8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$in equation (2).

$$\begin{gathered} E_1=\frac{2.1×{10}^{-5}C}{2×\mathrm{π}{\left(0.5\mathrm{m}\right)}^2×8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}·\frac{{10}^{-3}}{0.5\mathrm{m}} \\ =\frac{2.09×{10}^{-8}\mathrm{Cm}}{4.425×{10}^{-12}{\mathrm{C}}^2/N}·\frac{1}{0.5\mathrm{m}} \\ =3000N/C \end{gathered}$$

Thus, the maximum charge that can be placed on the disk is$2.09×{10}^{-5}C$ and the strength of the fringe field is$3000N/C$ .