FIGURE P27.42 shows a $4.0cm$wide plastic film being wrapped onto a $2.0cm$diameter roller that turns at $90rpm.$The plastic has a uniform surface charge density of $-2.0\mathrm{nC}/{\mathrm{cm}}^2$.

a. What is the current of the moving film?

b. How long does it take the roller to accumulate a charge of $-10\mu C?$?

Plastic film roller wide $=4.0cm$

Plastic film roller diameter$=2.0cm$

Number of rotations$=90rpm$

Uniform surface charge density$=-2.0\mathrm{nC}/{\mathrm{cm}}^2$

Consider a rotating cylinder with a frequency of $f$, which can be easily determined to be $1.5$.

The length L rounded per second will be

$L=\pi df$

Surface charge density be denoted by $\sigma =2\mathrm{nC}/{\mathrm{cm}}^2$,

Charge $Q$"intake" per second will be

$Q=\sigma aL=\sigma \pi adf$

Where $a=4\mathrm{cm}$is the side and the length units are all substituted in $\mathrm{cm}$.

Hence,

$Q=\left(2s\times \pi \times 4cm\times 2cm\times 1.5Hz\right)$

$Q=75.4\mathrm{nC}$

Hence, the current of the moving film is$75.4nC$

Plastic film roller wide$=4.0\mathrm{cm}$

Plastic film roller diameter$=2.0\mathrm{cm}$

Number of rotations$=90rpm$

Uniform surface charge density $=-2.0\mathrm{nC}/{\mathrm{cm}}^2$

Charge$=-10\mu C$

The time, in seconds, will be

$t=\left(\frac{InitialCharge}{Chargeinmovingfilm}\right)$

$t=\left(\frac{10\times {10}^{-6}C}{75.4\times {10}^{-9}C}\right)$

$t=133\mathrm{s}$

Hence, the time take will be $133s$for the roller to accumulate $-10\mu C$charge.