I understand A&B please show all the work for every other part to be awarded poi
ID: 2290741 • Letter: I
Question
I understand A&B please show all the work for every other part to be awarded points
A capacitor consisting of conducting
coaxial cylinders of radii a=1.0 cm and b=3.0 cm,
respectively, and length l=25.0 cm is connected to a
voltage source, as shown below. When the capacitor is
charged, the inner cylinder develops a charge of Q=+12?C.
Neglect end effects and assume that the region between the
cylinders is filled with air (?=1.0).
a) Where does the charge on the inner conductor
reside? Explain.
b) What charge is on the outer conductor? Explain.
c) Use Gauss's Law to determine an expression for the electric field at a distance r from the axis of the
cylinder where a < r < b. Show work and reasoning.
Answers :
a) On the surface
b) -12 ?C
c) 2.16x105
Nm/c * 1/r
d) ?V = -236 kV
e) 51 pF
f) E down by factor of 3, V drop by factor of 3, C increases by factor of 3
Explanation / Answer
"A cylindrical shell of length 190 m and radius 4 cm carries a uniform surface charge density of ? = 12 nC/m2.
(a) What is the total charge on the shell?
Find the electric field at the ends of the following radial distances from the long axis of the cylinder.
(b) r = 2 cm"
To part (a) I answered 573nC
circumference*length*(charge density)=(2pi*.04)*190*12E-9=573E-9
Part b is obviously very difficult to calculate absolutely correctly, since there is no obvious gaussian surface perfectly orthogonal to the electric field at every point. However, it seems to me that a short cylinder with its axis aligned with the charged cylindrical surface would make a good approximation, and I believe that is what is expected of me in solving this problem.
So I'll do it with a cylinder of length 1m for simplicity's sake. Since (to a good approximation) the field will be orthogonal to the cylinder's curved surface at all points, and the flux through the ends will be 0, the flux through the curved survace will equal the field strength times the surface area of this surface, which will also equal the charge enclosed over epsilon naught (natural permittivity of space, I'll write this e0).
E*A=Q/e0
E=Q/(e0*A)
So the charge enclosed will simply be the charge in a 1m section of the charged cylinder.
Q=(2pi*.04)*12E-9
The surface area of my Gaussian surface is...
A=(2pi*.02)
(ends not included because the flux through the ends of the cylinder is 0)
E=.04*12E-9/(en*.02)=2710.581760219553