Problem 1 In class, we calculated the potential due to a uniformly charged disk

Question

Problem 1 In class, we calculated the potential due to a uniformly charged disk with radius R and surface charge density , at a point z along the z-axis 2E0 Now let's model a parallel plate capacitor as a pair of disks separated by a distance d. One is situated at 2-d/2 with surface charge density , and the other at z- -d/2 with surface charge density -o. a) Write down V (z) for this charge distribution. Sketch a plot of V(z) from z =-R to z = R. For this plot, choose d- R/10 so that the plates are reasonably close together. Based on your plot, is the electric field between the plates roughly uniform? b) Now let's look at the limiting case where d 0, but oo in such a way that the product d is kept finite. Show that in this limit, the potential along the z-axis is 11 2 V(z) = sign(z) 20 where sign(z) is the sign of z, equal to +1 when z 0 and-1 when z

Explanation / Answer

given
V(z) = (sigma/2*epsilon)(sqroot(R^2 + z^2) - |z|)

a. for the given distribution
for for -R < z < 0
V(z) = (sigma/2*epsilon)(sqroot(R^2 + z^2) + z)

for 0 < z < R
V(z) = (sigma/2*epsilon)*(sqroot(R^2 + z^2) - z)

let d = R/10
then, from the plots
the electric potential between the plates is falling linearly
hecne the elctric field is fairly uniform

b. when d -> 0, but sigma -> inf such that sigma*d = finite
then V(z) = (sigma*d/2*epsilon)(sqroot((R/d)^2 + (z/d)^2) - z/d)
hecne
V(z) = (Product/2*epsilon)(sign(z) - z/sqrt(z^2 + R^2))
where sgn is signum fucntion

c. for R < < z
V(z) = (gigma/2*epsilon)z*(1 - sgn(z))

d. if we taylor expand
V(z) = (sigma/2*epsilon)*x(1 + 0.5(z/R)^2 - z)

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