The volume charge density in the tube wall is p. This tube has inner radius a, and outer radius b, and length L that is much greater than b. The region inside the tube (r < a, darker region,) and the region outside the tube (r > b) are empty. (The reason it is important that the tube is made of an insulator is that we can have charge distributed throughout its material; As we have discussed (or will soon,) all the excess charge on a conductor ends up at the surfaces of the conductor. Using Gauss's law, find the magnitude of the electric field at any point inside the tube wall (a < r < b) as a function of the radial distance r from the center. Follow these steps: (a) Choose a suitable Gaussian surface and indicate it in a diagram that you draw. Find an ex- pression for the charge enclosed by this Gaussian surface. Note that there charge is only in the walls of the tube! (b) Find an expression for the net electric flux through your chosen Gaussian surface. Clearly state and justify briefly any assumptions made. (c) Now use your results from (a) and (b) together with Gauss's law to find the magnitude of the electric field at a distance r from the center, where a < r < b. (d) Check that your answer makes sense, for example, at r = a and r = b. For r = a and with a concentric cylindrical surface as the chosen Gaussian surface, the enclosed net charge will be zero, so the field should also be zero . Does your expression derived in step (c) correctly model this behavior? Show that at r > b, the field should be the same as that of a long line of uniform charge whose charge per unit length is given by the charge density p multiplied by the cross sectional area of the tube wall. Does your expression derived in step (c) correctly model this behavior?
The volume charge density in the tube wall is p. This tube has inner radius a, and outer radius b, and length L that is much greater than b. The region inside the tube (r < a, darker region,) and the region outside the tube (r > b) are empty. (The reason it is important that the tube is made of an insulator is that we can have charge distributed throughout its material; As we have discussed (or will soon,) all the excess charge on a conductor ends up at the surfaces of the conductor. Using Gauss's law, find the magnitude of the electric field at any point inside the tube wall (a < r < b) as a function of the radial distance r from the center. Follow these steps: (a) Choose a suitable Gaussian surface and indicate it in a diagram that you draw. Find an ex- pression for the charge enclosed by this Gaussian surface. Note that there charge is only in the walls of the tube! (b) Find an expression for the net electric flux through your chosen Gaussian surface. Clearly state and justify briefly any assumptions made. (c) Now use your results from (a) and (b) together with Gauss's law to find the magnitude of the electric field at a distance r from the center, where a < r < b. (d) Check that your answer makes sense, for example, at r = a and r = b. For r = a and with a concentric cylindrical surface as the chosen Gaussian surface, the enclosed net charge will be zero, so the field should also be zero . Does your expression derived in step (c) correctly model this behavior? Show that at r > b, the field should be the same as that of a long line of uniform charge whose charge per unit length is given by the charge density p multiplied by the cross sectional area of the tube wall. Does your expression derived in step (c) correctly model this behavior?
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Answer all parts of the PHYSICS problem please
Transcribed Image Text:The volume charge density in the tube wall is p. This tube has inner radius a, and outer radius b,
and length L that is much greater than b. The region inside the tube (r < a, darker region,) and
the region outside the tube (r > b) are empty. (The reason it is important that the tube is made of
an insulator is that we can have charge distributed throughout its material; As we have discussed
(or will soon,) all the excess charge on a conductor ends up at the surfaces of the conductor.
Using Gauss's law, find the magnitude of the electric field at any point inside the tube wall (a < r < b)
as a function of the radial distance r from the center. Follow these steps:
(a) Choose a suitable Gaussian surface and indicate it in a diagram that you draw. Find an ex-
pression for the charge enclosed by this Gaussian surface. Note that there charge is only in the
walls of the tube!
(b) Find an expression for the net electric flux through your chosen Gaussian surface. Clearly state
and justify briefly any assumptions made.
(c) Now use your results from (a) and (b) together with Gauss's law to find the magnitude of the
electric field at a distance r from the center, where a <r < b.
b. For r = a and with a
(d) Check that your answer makes sense, for example, at r = a and r =
concentric cylindrical surface as the chosen Gaussian surface, the enclosed net charge will be
zero, so the field should also be zero . Does your expression derived in step (c) correctly model
this behavior? Show that at r > b, the field should be the same as that of a long line of uniform
charge whose charge per unit length is given by the charge density p multiplied by the cross
sectional area of the tube wall. Does your expression derived in step (c) correctly model this
behavior?
Transcribed Image Text:+ +
+
+ +
+
+
+
+
a
+
+
+
+
+
+
b + +
+
The figure above shows a cross section of a very long, hollow dielectric (insulating) tube (the outer,
lighter colored region with + signs) with charge distributed uniformly throughout the tube walls.
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