Electric field due to concentric cylinders

[stunner5000pt]

2 long charged concentric cylinders have radii of 3.22cm and 6.18cm. Surface charge densit of the inner cylinder is 2.41 micro C / m^2 and outer cylinger is -18.0microC/ m^2. Find electric field at r = 4.10cm (r is the radius as taken from the central axid of these two concetric cylinders)

i worked it out and got E = 2 k (2.41 x 10^-6)/ (0.041) = 1.06x 10^6N/C

am i right ?

the book gives an answer that is 2.12 MN/C

so who i right me or the book?

[Allday]

I got an answer of .214 M N/C, but whats a factor of ten between friends? When you use Gauss's Law you should have

E*2*pi*r*L, on one side and, (2*pi*a*L*sigma)/epsilon, on the other

L is any arbitrary length (it cancels) r is where you want the field, a is the radius of the inner cyl. sigma is the surface charge density and epsilon is epsilon.

Gabriel

[stunner5000pt]

I got an answer of .214 M N/C, but whats a factor of ten between friends? When you use Gauss's Law you should have

E*2*pi*r*L, on one side and, (2*pi*a*L*sigma)/epsilon, on the other

L is any arbitrary length (it cancels) r is where you want the field, a is the radius of the inner cyl. sigma is the surface charge density and epsilon is epsilon.

Gabriel

there must a reason why its off my a tenth though

[Allday]

I got

E = (a * sigma)/(r * epsilon)

a = .0322 m
sigma = 2.41 * 10^-6 C/m^2
r = .0410 m
epsilon = 8.85 * 10^-12

This gives me 213868 N/C

Gabriel

[stunner5000pt]

I got

E = (a * sigma)/(r * epsilon)

a = .0322 m
sigma = 2.41 * 10^-6 C/m^2
r = .0410 m
epsilon = 8.85 * 10^-12

This gives me 213868 N/C

Gabriel

I have to make a correction, the answer in teh book is 2.19 MN/C

so we're off by a tenth and and a few points

perhaps the solution you are posing is wrong??