Physics Worksheet With Answers - Terry Honan - Blinn College Page 4
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Chapter B - Problems
For r > R all af the charge is inside the Gaussian surface.
r = 15 cm ï Q
= Q = 216 mC
inside
(b) For any conductor in electrostatics, since the electric field is zero inside it all of the charge is at its surface.
r < R (r = 4 cm and r = 6 cm) ï Q
= 0
inside
As before, for r > R all af the charge is inside the Gaussian surface.
r = 15 cm ï Q
= Q = 216 mC
inside
Problem B.7
A 12 m long line with a uniform charge of 3 mC. Consider a tube of radius 8 cm and of length 2 cm; adding 8 cm radius disks to each
end makes it a closed surface. Take this to be our Gaussian surface. Take the line to be the central axis of the tube and the center of
the line to be the center of the tube.
(a) What is the total charge enclosed by the Gaussian surface?
(b) Using the assumptions that 2 cm and 8 cm are both much less than 12 m, find the electric field at the surface of the tube.
Solution to B.7
-6
-7
(a) The charge per length of the line of charge is l = 3 µ 10
ë 12 = 2.5 µ 10
C ê m. The charge per length times the length of the
Gaussian surface gives Q
.
inside
-9
= l µ 0.02 m = 5 µ 10
Q
C
inside
(b) Making the assumptions that the length of the line charge is much longer than the other lengths causes the field to point radially
away from the line. At every point on the tube the field is perpendicular to the tube. We get
E × „ A = E µ A
= E µ 2 p r L.
Ÿ
tube
tube
The flux through the entire Gaussian surface is the same as the flux through the tube. The field is parallel to the face of the disks at
the ends and thus the flux through them is zero.
E × „ A = E µ 2 p r L
ò
Putting this together using Gauss's law gives the electric field magnitude.
-9
Q
5µ10
N
inside
ò
E × „ A =
ï E µ 2 p 0.08 µ 0.02 =
ï E = 56 200
-12
ε
C
8.85µ10
0
`
The direction of the field points radially away from the line. Call r
the unit vector in this direction. The field then becomes:
`
N
E = 56 200
r
.
C
Problem B.8
Consider the general case of spherical symmetry where there is a charge distribution given by rHrL, where r is the radial distance from
some origin.
(a) Find a general expression for the electric field as a function of postion in terms of an integral over the charge density.
(b) For the case of rHrL = a ê r find the field as a function of position.
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