A long cylindrical insulator has a uniform charge density of 1.0 #C/m3 and a radius of
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?
2 answers
# means micro
Use 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.
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.
1 answer