91Ó°ÊÓ

The given data can be listed below,

• The radius of plastic disk is,$\mathrm{R}=1.5 \mathrm{m}$
• The charge on the plastic disk is,$-\mathrm{Q}=-3×{10}^{-5} \mathrm{C}$
• The distance between plastic disk and aluminium foil is,$\mathrm{d}=3 \mathrm{mm}\left(\frac{{10}^{-3} \mathrm{m}}{1\mathrm{mm}}\right)$
• The radius of aluminium foil is,role="math" localid="1656936726553" $\mathrm{r}=2 \mathrm{cm}\left(\frac{1 \mathrm{m}}{100\mathrm{cm}}\right)$
• The thickness of aluminium foil is,$\mathrm{t}=1 \mathrm{mm}\left(\frac{{10}^{-3} \mathrm{m}}{1 \mathrm{mm}}\right)$

The collection or redistribution of electric charges in a body as a result of a nearby charged body without any physical contact is known as electrostatic induction.

The charge induced on the right side of aluminium foil is negative and on the left side it is positive to keep it neutral. A induced charge on the surface of conductor can cancel out the electric field inside it. So, the charge distribution is shown below,

Thus, due to the induction of charges right side of the foil acquire a negative charge, and the left side of the foil acquires a positive charge.

The aluminium foil is a conductor, which implies that no electric field can exist on the inside, resulting in a zero field at the location. Until all external electric fields are cancelled, a conductor has enough free charge to travel to the surface.

Thus, the electric field at the center of foil is zero.

The electric field of the plastic ring at the center of the aluminum disk is given by,

$$\begin{gathered} {\mathrm{E}}_p=\frac{\left(-Q/A\right)}{2{\mathrm{Î\mu }}_0}\left[1-\frac{\mathrm{d}+\mathrm{t}/2}{{\left({\mathrm{R}}^2+{\left(\mathrm{d}+\mathrm{t}/2\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ ≈\frac{\left(-Q/A\right)}{2{\mathrm{Î\mu }}_0} \end{gathered}$$

The electric field of the aluminum disk's surfaces are given by,

$$\begin{gathered} {\mathrm{E}}_{+}=\frac{\left(-\mathrm{q}/\mathrm{a}\right)}{2{\mathrm{Î\mu }}_0}\left[1-\frac{\mathrm{t}/2}{{\left({\mathrm{R}}^2+{\left(\mathrm{t}/2\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ ≈\frac{\left(\mathrm{q}/\mathrm{a}\right)}{2{\mathrm{Î\mu }}_0} \end{gathered}$$

And,

role="math" localid="1656994121348" $$\begin{gathered} {\mathrm{E}}_{-}=\frac{\left(-\mathrm{q}/\mathrm{a}\right)}{2{\mathrm{Î\mu }}_0}\left[1-\frac{\mathrm{t}/2}{{\left({\mathrm{R}}^2+{\left(\mathrm{t}/2\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ ≈\frac{\left(\mathrm{q}/\mathrm{a}\right)}{2{\mathrm{Î\mu }}_0} \end{gathered}$$

Here, q is the induced charge and a is the area of aluminium foil.

The net electric field inside the foil is given by,

${\mathrm{E}}_{+}+{\mathrm{E}}_{-}+{\mathrm{E}}_{\mathrm{P}}=0$

Substitute all the values in the above expression the charge on foil is given by,

$$\begin{gathered} \frac{\left(-\mathrm{Q}/\mathrm{A}\right)}{2{\mathrm{Î\mu }}_0}+\frac{\left(\mathrm{q}/\mathrm{a}\right)}{2{\mathrm{Î\mu }}_0}+\frac{\left(\mathrm{q}/\mathrm{a}\right)}{2{\mathrm{Î\mu }}_0}=0 \\ 2\frac{\mathrm{q}}{\mathrm{a}}=\frac{\mathrm{Q}}{\mathrm{A}} \\ \mathrm{q}=\frac{\mathrm{Qa}}{2\mathrm{A}} \end{gathered}$$

Substitute values in the above,

$$\begin{gathered} \mathrm{q}=\frac{\mathrm{Q}}{2}\left(\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right) \\ =\frac{3×{10}^{-5}\mathrm{C}}{2}{\left(\frac{0.02\mathrm{m}}{1.5\mathrm{m}}\right)}^2 \\ =2.67×{10}^{-9}\mathrm{C} \end{gathered}$$

Thus, the charge on left circular face of aluminum foil is$2.67×{10}^{-9}\mathrm{C}$ .