(a) The drum of a photocopying machine has a length of 42 cmand a diameter of 12 cm.The electric field just above the drum’s surface is $\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}\mathbf{}}\mathbf{N}\mathbf{/}\mathbf{C}$ .What is the total charge on the drum? (b) The manufacturer wishes to produce a desktop version of the machine. This requires reducing the drum length to 28.0 cmand the diameter to 8.0 cm.The electric field at the drum surface must not change. What must be the charge on this new drum?

Case-1

1. Length of the drum,$\mathrm{I}=42\mathrm{cm}$
2. Diameter of the drum,$\mathrm{d}=12\mathrm{cm}$
3. Electric field on the drum’s surface,$\mathrm{E}=2.3\times {10}^5\mathrm{N}/\mathrm{C}$

Case-2

1. Length of the new drum, $\mathrm{I}\text{'}=28\mathrm{cm}$
2. Diameter of the new drum,$\mathrm{d}\text{'}=8\mathrm{cm}$
3. The electric field does not change.

Using the concept of the electric flux of Gauss's law, we can get the expression of the charge of the particle. Hence, by substituting the given data, we can get the charge on the respective drums.

Formula:

The electric charge enclosed within the Gaussian surface,

${\mathrm{q}}_{\mathrm{en}\mathrm{c}}={\mathrm{\epsilon }}_0\mathrm{\varphi }=2{\mathrm{\pi r\epsilon }}_0\mathrm{LE}$ (1)

Using the given data in equation (1), we can get the charge of the drum as follows:

( for 2r=D)

$$\begin{gathered} \mathrm{q}={\mathrm{\pi \epsilon }}_0\mathrm{EDh} \\ =\left(8.85\times {10}^{12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(2.3\times {10}^5\mathrm{N}/\mathrm{C}\right)\left(0.12\mathrm{m}\right)\left(0.42\mathrm{m}\right) \\ =3.2\times {10}^{-7}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $3.2\times {10}^{-7}\mathrm{C}$.

Using the given data in equation (i) and comparing the charge to the old drum, we can get the charge of the drum as follows:

(for 2r=D )

$$\begin{gathered} \mathrm{q}\text{'}=\mathrm{q}\left(\frac{\mathrm{A}\text{'}}{\mathrm{A}}\right) \\ =\mathrm{q}\left(\frac{\mathrm{\pi D}\text{'}\mathrm{h}\text{'}}{\mathrm{\pi Dh}}\right) \\ =\left(3.2\times {10}^{-7}\mathrm{C}\right)\left(\frac{\left(8.0\mathrm{cm}\right)\left(28\mathrm{cm}\right)}{\left(12\mathrm{cm}\right)\left(42\mathrm{cm}\right)}\right) \\ =1.4\times {10}^{-7}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $1.4\times {10}^{-7}\mathrm{C}$.