A very long solid cylinder of radius R=3 cm has a uniform volume charge density
ID: 1430445 • Letter: A
Question
A very long solid cylinder of radius R=3 cm has a uniform volume charge density p =+6 mue c/m^3. How much total charge is contained on a 1m length of this cylinder? What would be the effective linear charge density along the length of this cylinder? For a given length of cylinder, what percentage of its total charge lies between 0 and R/2? What is the electric field at a radial distance of 4 cm from the axis is the cylinder? What is the electric field at a radial distance of 2 cm from the axis of the cylinder? Now sketch E_r vs r from r=0 to infinity. Mark off where r= R occurs. Inside the cylinder (rR), graph's shape is Setting V-0 at r=3R, find the voltage at the following locations: V(r=0)= V(=R/2)= V(r=R)= V(r=2R)= V(r=4R)= Now sketch v vs r from r=0 to infinity. Mark off where r=R occurs. Inside the cylinder (rR), graph's shape isExplanation / Answer
a)
i) cross section area of cylinder = pi R^2 = pi ( 0.03^2)
A = 2.827 x 10^-3 m^2
volume of 1m length part of cylinder = AL = 2.827 x 10^-3 m^3
Charge = charge density x volume
= 2.827 x10^-3 x 6 x 10^-6 = 1.70 x 10^-8 C
ii) linear charge density = charge volume density x cross section area
= 6 x 10^-6 C/m^3 x 2.827 x 10^-3 m^2
= 1.70 x 10^-8 C / m
iii) DENSITY is same everywhere .
so charge fraction will be same as cross section area fraction.
A_1/2 = pi (R/2)^2 = pi R^2 / 4
A = pi R^2
Q = [(pi R^2/ 4) / (pi R^2)] x 100
Q = 25 %
b) using Gauss law,
flux = E.A = Qinside / e0
E ( 2 pi r L ) = ( 1.70 x 10^-8 L ) / e0
E = (1.70 x 10^-8 ) / (2 x pi x 0.04 x 8.854 x 10^-12)
E = 7623.67 N/C
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at r = 0.02 m
E ( 2 pi r L ) = ( 6 x 10^-6 x r^2 / R^2 ) / e0
E (2 pi ) = (6 x 10^-6 x r) / (e0 R^2)
E = (6 x 10^-6 x 0.02 ) / (2 x pi x 0.03^2 x 8.854 x 10^-12)
E = 2.40 x 10^6 N/C
E (r < R) = (rho * r) / (2pi R^2 e0 )
E = a r
where a is constant
graph will straight line with slope a.
ANs. B
for E (r > R)
E = (rho * A ) / (2 pi e0 r)
E = a / r
Ans. D