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December 21, 2014

December 21, 2014

Posted by **Fatima** on Saturday, March 10, 2012 at 12:08pm.

9.0 cm.

a) What is the electric field inside the insulator at a distance of 4 cm?

b) What is the electric field at 17 cm?

c) How much work must you do to bring a q = 0.05 #C test charge from 17 cm

to 4 cm?

- Physics -
**Fatima**, Saturday, March 10, 2012 at 12:09pm# means micro

- Physics -
**drwls**, Saturday, March 10, 2012 at 6:34pmUse Gauss' Law inside and outside the cylinder

a) Inside a uniformly charged cylinder of radius r < 0.09 m, there will be

Q = 10^-6 * pi*r^2*L Coulombs

Gauss' Law says, for this geometry, that

2*pi*r*L*E(r) = Q/epsilon

= 10^-6 * pi*r^2*L

Therefore

E(r) = 10^-6*r/(2*epsilon)

= 2260 Volt/m at r = 0.04 m

For epsilon, I used 8.85*10^-12 C^2/(N*m^2)

(b) Let R = 0.09 m be the solid cylinder radius. r = 17 cm is outside the solid cylinder. The charge inside an outer cylinder of radius r > R is

Q = 10^-6*pi*R^2*L

and the E- field at r > R is again given by Gauss' law

2 pi*r*L*E(r) = 10^-6*pi*R^2*L

E(r) = 10^-6 R^2/(2*r*epsilon)

= 5085*(R/r) Volt/m

c) integrate

(test charge)*E*dr from r = 17 to r = 4 cm. You will have to do in in two parts, inside and outside r = R, since the formulas are different in the two regions.

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