91Ó°ÊÓ

The emf of the flashlight battery is $\mathrm{Î\mu }=1.7\text{V}$

The length of battery is $l=4.5\text{cm}$

The radius of cylindrical battery is$r=1\text{cm}$

The amount of charge at the positive end of battery plate is found by the product of the capacitance and emf of the battery.

The capacitance of the cylindrical battery is given as:$C=\frac{{\mathrm{Î\mu }}_0\left(\mathrm{Ï€}{r}^2\right)}{l}$

Here, ${\mathrm{Î\mu }}_0$is the permeability of free space and its value is$8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2$ .

Substitute all the values in above equation.

$$\begin{gathered} C=\frac{\left(8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2\right)\left[\mathrm{π}{\left(\left(1\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right]}{\left(4.5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)} \\ C=6.18×{10}^{-14}\text{F} \end{gathered}$$

The amount of charge on the positive end of the battery plate is given as:

$q=C\mathrm{Î\mu }$

Substitute all the values in above equation.

$$\begin{gathered} q=\left(6.18×{10}^{-14}\text{F}\right)\left(1.7\text{V}\right) \\ q=1.051×{10}^{-13}\text{C} \end{gathered}$$

Therefore, the amount of charge on the positive end of the battery plate is$1.051×{10}^{-13}\text{C}$