When a shower is turned on in a closed bathroom, the splashing of the water on t

Question

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 2.3 m x 4.3 m x 2.2 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 601 N/C. Also, treat those surfaces as forming a closed Gaussian surface around the room's air. What are (a) the volume charge density and (b) the number of excess elementary charges e per cubic meter in the room's air?

Explanation / Answer

given dimensions of bath room 2.3m X 4.3 m X 2.2 m

   electric field E = 601 N/C

we can calculate total surface area of the bathroom as A = 2(2.3*4.3)+2(4.3*2.2)+2(2.2*2.3)= 48.82 m2

we know that by definition of electric flux as phi = E*A cos theta ( theta is zero here, given conditions)
              

           phi = 601*48.82 = 29340.82 Nm2/C

   From Gauss' law electric flux = 1/ epsilon not (charge inside)

               q = phi*epsilon not
               q = 29340.82*8.854*10^-12 C
               q = - 2.6*10^-7 C
now volume of the bath room is v = 2.3m X 4.3 m X 2.2 m = 21.758 m3

charge density rho = q/ v = - 2.6*10^-7/21.758 = -1.2*10^-8 C/m3

b) the number of excess elementary charges e per cubic meter in the room's air is

       N =(q/elementary charge)/volume
       N = (2.6*10^-7/1.6*10^-19)/21.758
N = 74685173269.60
       N = 7.46852*10^10

the number of excess elementary charges e per cubic meter in the room's air is = 7.46852*10^10

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