When a shower is turned on in a closed bathroom, the splashing of the water on the bare tub can fill the room’s air with negatively charged ions and produce an electric field in the air as great as $\mathbf{1000}\mathbf{N}\mathbf{/}\mathbf{C}$. Consider a bathroom with dimensions$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{m}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}$. Along the ceiling, floor, and four walls, approximate the electric field in the air as being directed perpendicular to the surface and as having a uniform magnitude of$\mathbf{600}\mathbf{N}\mathbf{/}\mathbf{C}$. Also, treat those surfaces as forming a closed Gaussian surface around the room’s air. What are (a) the volume charge density r and (b) the number of excess elementary charges eper cubic meter in the room’s air?

1. The electric field in the air,${\mathrm{E}}_{\mathrm{air}}=1000\mathrm{N}/\mathrm{C}$
2. Dimensions of the bathroomrole="math" localid="1657343746215" $2.5\mathrm{m}\times 3.0\mathrm{m}\times 2.0\mathrm{m}$
3. Uniform magnitude of the electric field perpendicular to the surface:${\mathrm{E}}_{\mathrm{room}}=600\mathrm{N}/\mathrm{C}$

Using the concept of the Gauss flux theorem, we can get the net flux through the surface. Now, using this value, we can get the enclosed charge. This value divided by the total volume will give the volume charge density. Now, using the charge value divided by the charge of the electron, we can get the required number of excess electrons.

Formulae:

The electric flux passing through the surface,

$\mathrm{\varphi }=\mathrm{E}.\mathrm{A}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (1)

The volume charge density,

$\mathrm{\rho }=\frac{\mathrm{q}}{\mathrm{V}}$ (2)

The number of excess electrons,

$\mathrm{n}=\frac{\mathrm{q}}{\mathrm{e}}$ (3)

The total surface area bounding the bathroom using its dimensions is given by,

$$\begin{gathered} \mathrm{A}=2\left(2.5\mathrm{m}\times 3.0\mathrm{m}\right)+2\left(3.0\mathrm{m}\times 2.0\mathrm{m}\right)+2\left(2.0\mathrm{m}\times 2.5\mathrm{m}\right) \\ =37{\mathrm{m}}^2 \end{gathered}$$

The absolute value of the total electric flux, with the assumptions stated in the problem in the equation (1) is given as:

$\left|\mathrm{\varphi }\right|=(600\mathrm{N}/\mathrm{C})\left(37{\mathrm{m}}^2\right)$

By Gauss’ law, we conclude that the enclosed charge (in absolute value) is given using equation (i) as:

$$\begin{gathered} \left|{\mathrm{q}}_{\mathrm{enc}}\right|=\left(8.85\times {10}^{-12}{\mathrm{C}}^2.{\mathrm{m}}^2/\mathrm{N}\right)\left(600\mathrm{N}/\mathrm{C}\right)\left(37{\mathrm{m}}^2\right) \\ =2.0\times {10}^{-7}\mathrm{C} \end{gathered}$$.

Therefore, with volume and recognizing that we are dealing with negative charges, the volume charge density is given using equation (2), such that,

$$\begin{gathered} \mathrm{\rho }=\frac{-2.0\times {10}^{-7}\mathrm{C}}{15{\mathrm{m}}^3} \\ =-1.3\times {10}^{-8}\mathrm{C}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of the charge density is $-1.3\times {10}^{-8}\mathrm{C}/{\mathrm{m}}^3$.

The number of excess electrons per cubic meter is found by dividing equation (3) by volume as follows:

$$\begin{gathered} \mathrm{n}/\mathrm{V}=\left(2.0\times {10}^{-7}\mathrm{C}/1.6\times {10}^{-19}\right)/(15{\mathrm{m}}^3) \\ =8.2\times {10}^{10}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the required value of electrons per cubic meter is $8.2\times {10}^{10}/{\mathrm{m}}^3$.