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 1000 N/C. Consider a bathroom with dimensions 2.5 m x 3.0 m x 2.0 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 600 N/C. Also, treat those surfaces as forming a closed Gaussian surface around the room’s air. What are (a) the volume charge density \rho and (b) the number of excess elementary charges e per cubic meter in the room’s air?

#### Solution:

Using Gauss’s law, we can calculate the total flux. Gauss’s law states:

\Phi=(E)(A)(Where \Phi is flux, E is the electric field, and A is the area)

The total flux is = \Phi=(37)(600)=22200 \dfrac{N\cdot m^2}{C}

To find the density, we first need to find the charge enclosed. Again, we can use Gauss’s law to figure it out. Gauss’s law states:

(Where \Phi is the net flux, \epsilon_0 is permittivity of free space and q is the charge enclosed)

q=(\Phi)(\epsilon_0)

Substitute the values we know:

q=(22200)(8.85\times 10^{-12})=1.96\times10^{-7} C

a) To find the density, we need to calculate the volume of the room and divide the charge enclosed by the volume.

\rho=-1.3\times10^{-8}\dfrac{C}{m^3}

(Our answer is negative because the room is filled with negatively charged ions)

b) Let us now calculate the excess elementary charges e per cubic meter.

Thus, there are 8.167\times10^{10} excess elementary charges per cubic meter.

(Note that 1.6\times10^{-19} is the charge of an electron)

#### Final Answers:

There are 8.167\times10^{10} excess elementary charges per cubic meter.