Question

2. Suppose the hw 1 rat cage floor was 0.30m * 0.60m with ceiling 0.20m above it. a) What charge on the ceiling would fill the cage with 180 N/c () field intensity? b) the floor should have { zero | same opposite half double } the ceiling charge.

Answer 1

2. Given,
rat cage floor area is Af = 0.3*0.6 = 0.18 m^2
height of the cage is h = 0.2 m

a. let charge on ceiling be Q
this charge will fill cage will electric field E = 180 N/C fro mbottom to top
for this the charge has to be -ve (as it is on the roof of the cage)
also,
Q/Af = 2E*epsilon0
Q = Af*2E*epsilon0 = 0.18*2*180*8.854*10^-12
Q = 573.7392*10^-12 C
  
b. for the Q we calculated in part A. the floor needs to have 0 charge
but, if we assume that net charge on the cage is 0
then we can see that electric field does not depend on hte distance from the sheet, but only on the charge density
hence, if net charge on cage is non 0, then bottom floor has 0 charge
if net charge on the cage is 0, then the floor has equal charge to the top floor opposite in sign (=ve in nature) with value of new Q' = Q/2 = 286.8696*10^-12 C