Question

A$2.0\mathrm{m}\times 4.0\mathrm{m}$flat carpet acquires a uniformly distributed charge of $-10\mathrm{\mu C}$after you and your friends walk across it several times. A$2.5\mathrm{\mu g}$dust particle is suspended in midair just above the center of the carpet. What is the charge on the dust particle?

Short answer

The charge on the dust particle is$-3.47\times {10}^{-13}\mathrm{C}$.

Step-by-step solution

Calculation of charge density

Given,

$\mathrm{Q}=-10\times {10}^{-6}\mathrm{C}$

$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

$\mathrm{m}=0.0025\times {10}^{-6}\mathrm{kg}$

${\mathrm{ϵ}}_{\mathrm{o}}=8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$

$\mathrm{F}=\mathrm{mg}=\mathrm{qE}$

$\mathrm{\eta }=\frac{\mathrm{Q}}{\mathrm{A}}$

The charge density by dividing the charge$\left(\mathrm{Q}\right)$by the area$(2.0\times 4.0)\mathrm{m}$,

$\mathrm{\eta }=\frac{-10\times {10}^{-6}\mathrm{C}}{(2.0\mathrm{m}\times 4.0\mathrm{m})}$

$\mathrm{\eta }=\frac{-1\times {10}^{-5}}{8.0\mathrm{m}}$

$\mathrm{\eta }=-1.25\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2$

Calculation of Electric field 

To calculate the electric field, multiply the charge density by the temporal permittivity constant.

Electric field,

$\mathrm{E}=\frac{\mathrm{\eta }}{2{\mathrm{ϵ}}_{\mathrm{o}}}$

$\mathrm{E}=\frac{-1.25\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2}{\left(2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}$

$\mathrm{E}=-7.06\times {10}^4\mathrm{N}/\mathrm{C}$

Calculation for charge

To calculate the charge of a dust particle, multiply the mass times gravity by the electrical field.

$\mathrm{q}=\frac{\mathrm{mg}}{\mathrm{E}}$

$\mathrm{q}=\frac{\left(0.0025\times {10}^{-6}\times 9.8\right)}{-7.3\times {10}^4}$

$\mathrm{q}=-3.47\times {10}^{-13}\mathrm{C}$