How Does Distance Affect Electric Field Calculation Above a Charged Sheet?

[gnarkil]

Homework Statement

what is the approx electric field 1cm above a typical sheet of paper with a surface charge density of sigma = 45 nC/m^2?

Homework Equations

electric field E = sigma/epsilon_0 where epsilon_0 = 8.85*10^-12

E = kq/r^2 where k = 9*10^9, r is distance in meters

assume paper is charged sheet, then E = sigma/2*epsilon_0

The Attempt at a Solution

i assumed that 1 cm is so small that the distance is insignificant and could be ignored, i think this is why i got it wrong. anyways...

E = sigma/epsilon_0 = (45*10^-8)/(8.85*10^-12) = 50847 N/C

it is wrong, how do i factor in the distance? should i have used the charged sheet formula instead or the general equation (the first one)?

 

[nasu]

Suggestion:
They say "a typical sheet of paper". They may want you to integrate over some typical dimensions (11"x8.5") rather than assume an infinite sheet.
How does the answer compares with your result? (greater, smaller)

 

[gnarkil]

that may be the case, but it wouldn't make sense to state a problem with such general guidelines, i don't think the problem involves specific dimensions

 

[scrplyr]

The equation for the field of a charged sheet is E= sigma/2*epsilon₀. You had the correct equation under "relevant equations" but i think you forgot to multiply epsilon by 2 in your calculations. Also check out section 21.5 in the book (page 358-359) The book states "with an infinite plane, symmetry requires that the field lines be perpendicular to the plane. So they don't spread out, and that means the field desnt vary with distance." which is why the equation doesn't include distance.

 

[gnarkil]

scrplyr, i tried what you suggested and it was still incorrect

E = sigma/2*epsilon_0 = (45*10^-8)/((2)(8.85*10^-12)) = 25423.728 N/C

what now?

 

[scrplyr]

45 nC/m^2 = 45*10^-9 C/m^2, not 10^-8

 

[gnarkil]

thanks, i got it correct, it was the frickin exponent!