A kitchen floor, which is 3.0 meters long by 4.0 meters wide, has a charge Q_xy

Question

A kitchen floor, which is 3.0 meters long by 4.0 meters wide, has a charge Q_xy =-2.0 mu C spread uniformly over the whole area. A point charge q_1 =- 5.0 mu C is held a distance 10 centimeters above the floor, near the middle of the room. What electric force acts on the point charge? (Derive any formulas you used, don't just plug in pre-derived text book equations. To use Gauss's law, take advantage of the fact that the floor is very long and wide compared to the distance to the point charge) When released, the point charge just floats in place. What Is its mass?

Explanation / Answer

a)
from the given data

surface charge density of the floor, sigma = Qo/Area

= -2*10^-6/(2*4)

= 0.167 micro C/m^2

now imagine a cyllnder with base area A which cuts the floor.

now use Gauss law,

net elctric flux through the gaussian cyllinder = Q_enclosed/epsilon

elctric flux flux through lower and upper surface of the cyllinder = sigma*A/epsilon

2*E*A = sigma*A/epsilon

E = sigma/(2*epsilon)

so, Force on q1, F = q1*E

= q1*sigma/(2*epsilon)

= 5*10^-6*0.167*10^-6/(2*8.854*10^-12)

= 0.047 N

b) in the equilibrium, Fg = Fe

m*g = Fe

m = Fe/g

= 0.047/9.8

= 0.0048 kg or 4.8 grams

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