ATMO/ECE 489/589 Spring 2017
Homework #1 Answers
1.
In my mind the power dissipated in the global electric circuit is roughly that generated by Tucson Electric's Irvington Rd. facility. That came as a bit of a surprise.
There are roughly 2000 thunderstorms active at any time; each is generating about 1 A of current. If we could somehow capture this energy it would replace just one medium size city's power generation plant.
2.
Here's a sketch of the situation we're dealing with.
Because 1500 m is much smaller than the radius of the earth we can consider the ground to be flat and E has just a z-component. In this problem we will use the differential form of Gauss' law. We assume that the volume space charge is distributed uniformly throughout the 1500 m thick layer.
Here are the details of the calculation.
You need to be careful with + and -
signs in a problem like this.
Another visual representation of
what is going on in this problem.
3.
In class we worked a problem where λo(Coulombs/meter) was distributed along an infinitesimally thin line. In this problem the same amount of charge per unit length is uniformly distributed in a cylinder of radius R. A length L of the line contains L λo Coulombs of charge. A length L of the cylinder contains the same L λo Coulombs of charge.
The fact that we're given both a line charge density ( λo ) and a volume space charge density ( ρ ) sometimes causes confusion. The relationship between λo and ρ is shown above.
We'll first find an expression for r < R.
We use the integral form of Gauss' Law. At any particular r the cylinder used in the area integral encloses just a portion of the charge in the cylinder. The expression Er is written in terms of both λo and ρ o.
Next we'll determine the field outside R.
3.
In class we worked a problem where λo(Coulombs/meter) was distributed along an infinitesimally thin line. In this problem the same amount of charge per unit length is uniformly distributed in a cylinder of radius R. A length L of the line contains L λo Coulombs of charge. A length L of the cylinder contains the same L λo Coulombs of charge.
The fact that we're given both a line charge density ( λo ) and a volume space charge density ( ρ ) sometimes causes confusion. The relationship between λo and ρ is shown above.
We'll first find an expression for r < R.
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We use the integral form of Gauss' Law. At any particular r the cylinder used in the area integral encloses just a portion of the charge in the cylinder. The expression Er is written in terms of both λo and ρ o.
Next we'll determine the field outside R.
Note this is the same
solution that we got when we determined
the field around an infinite uniform line
of charge (at r >> R a
cylinder of charge will be
indistinguishable from a line of charge).
Note that the two solutions ( r < R
) and ( r > R ) agree at r = R.
Finally we substitute in the value of breakdown voltage and determine the value of R.
Finally we substitute in the value of breakdown voltage and determine the value of R.