Question

(II) When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If that charge is\({\bf{3}}{\bf{.0}}\;{\bf{\mu C}}\), by what percentage does the mass of a 9.0-g comb change during charging?

Short answer

The mass of the comb is changed by \(1.9 \times {10^{ - 13}}\% \).

Step-by-step solution

Understanding the charging by rubbing

Two insulating materials may become electrically charged when they are rubbed against each other. In this process, electrons from one material migrate to another material. One material becomes negatively charged while another becomes positively charged.

Given Data

The charge on the comb is, \(q = 3\;{\rm{\mu C}} = 3 \times {10^{ - 6}}\;{\rm{C}}\).

The mass of comb is, \(m = 9\;{\rm{g}} = 9 \times {10^{ - 3}}\;{\rm{kg}}\).

Determination of the number of electrons moved

The number of electrons is given as:

\(n = \frac{q}{e}\)

Here, e is the electronic charge.

Substitute the values in the above expression.

\(\begin{aligned}{c}n = \frac{{3 \times {{10}^{ - 6}}\;{\rm{C}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C}}}}\\n = 18.75 \times {10^{12}}\end{aligned}\)

Determination of percentage of change in mass

The expression for the fractional change in the mass is given as:

\(\frac{{\Delta m}}{m} = \frac{{n{m_e}}}{m}\)

Here, \({m_e}\) is the mass of the electron.

Substitute the values in the above expression.

\(\begin{aligned}{l}\frac{{\Delta m}}{m} = \frac{{18.75 \times {{10}^{12}} \times 9.1 \times {{10}^{ - 31}}\;{\rm{kg}}}}{{9 \times {{10}^{ - 3}}\;{\rm{kg}}}}\\\frac{{\Delta m}}{m} = 18.958 \times {10^{ - 16}}\end{aligned}\)

The percentage change in the mass is,

\(\begin{aligned}{l}\frac{{\Delta m}}{m}\% = 18.958 \times {10^{ - 16}} \times 100\% \\\frac{{\Delta m}}{m}\% \approx 1.9 \times {10^{ - 13}}\% \end{aligned}\)

Thus, the percentage change in the mass of the comb is \(1.9 \times {10^{ - 13}}\% \).