During the Spring 2013 version of this course, the class before the Midterm Exam was devoted to review.

The exam will be open notes, open homework and homework answers. 

A copy of the first page of the Spring 2011 Midterm Exam was handed out in class (it has been pasted in below).  The format of this semester's exam will be very similar if not identical.  I.e. there will be a first problem consisting of several short answer questions.  You will probably have to choose 3 or 4 questions out of a larger group of questions.  You should probably plan on spending not more than 5 minutes on each answer.

Following that you will probably have to answer perhaps 3 longer questions that involve calculations more like you have been doing on the homework.  Again you will be able to choose which questions you answers from a little bit larger group.

The course this year is a little different than in 2013.  The class periods this semester are only 50 minutes long, they were 75 minutes long in 2013.  I will take that into consideration when designing this semester's exam.
Here are the first questions from the Spring 2011 exam:

ATMO/ECE 489/589 Midterm Exam

March 8, 2011

 

Everyone should answer Question #1.

ATMO/ECE 489 students should then answer any 3 of the remaining 5 questions

ATMO/ECE 589 students should then answer any 4 of the remaining 5 questions

Allow about 15 minutes per question

 

1.         Answer any three of the following questions.

(a)        Briefly explain what makes the atmosphere a weak conductor of electricity.  How and why does conductivity change with altitude?  Do electrical currents travel through air in the same way that current travels through a piece of wire?

 (b)       It seems curious to me that electrical charge can be produced in the cold wet environment inside a thunderstorm.  How is electrical charge created and distributed inside a thunderstorm?

(c)        List and discuss some of the contributions Benjamin Franklin made to the study of lightning and atmospheric electricity

(d)       Briefly describe how an electric field mill is able to measure static or slowly-varying electric fields.

(e)        Briefly explain the difference between “fast” and “slow” electric field change antenna systems.  Why do the signals from these antennas need to be integrated?  What is the difference between passive and active integrators?

 

2.         Thunderstorm fields just a few hundred meters above the ground are often much more intense than the fields at the ground.  In the figure below the field at 200 m reaches a peak amplitude of about 25 kV/m just before a lightning discharge.  By contrast corona discharge from trees, bushes, and other sharp objects on the ground added space to the air and limited the field at the ground to a value of about 8 kV/m.

           

 

 

 (a)        Use the peak field values at the ground and aloft to estimate the volume space charge density in the layer between the ground and 200 m altitude.  What is the polarity of this space charge?  (note: a positive electric field value means E is pointing upward)

 

(b)        Why is the value of the field change caused by the lightning discharge the same at the ground and at 200 m altitude (about 10 kV/m), even though the peak field values at the two altitudes are very different?


The first half of the 2013 Topics Summary is reproduced below (this summary was actually the one handed out in Spring 2011, the dates listed won't match up with class dates this semester.  You can download a similar summary prepared for the 2015 class.


Here are some comments made during the review held in 2013
1.  You should know typical values for the fair weather E field and atmospheric conductivity at ground level (100 - 300 V/m and 1 - 2 x 10-14 mhos/m, respectively).  A typical fair weather current density value, Jm, would then be 200 V/m x (1 x 10-14mhos/m) = 2 pA/m2

Somehow or another I forgot to include the lecture on Benjamin Franklin.  You should quickly skim that section. 

Short answer question example #1  
Question 1(c) on the Spring 2011 Midterm asks you to list and discuss some of the contributions made by Benjamin Franklin to the field of lightning and atmospheric electricity.  A list might include:

You could then say a little bit more about each of these.

More comments about items on the topics summary
2.  I would suggest having the values and units for constants together on one page in your notes so that you can find them quickly when needed on an exam problem.

3.  The two forms of Gauss' Law have been used a lot in the homework.  It's a near certainty that you will need to use them somewhere on the Midterm Exam.

4 & 5.  I would guess I'm much more likely to ask a problem that involves the electrostatic potential than I am to ask you to solve Laplace's or Poisson's equations (I don't think we have used Poisson's equation at all in the class up to this point)

6.  Charges inside a conductor will move around on the surface until the E field inside is zero.  You can use Gauss' Law to derive this very handy relationship between the surface charge density and the E field perpendicular to the conductor.

7.  There were at least a couple of homework questions that involved a charge (or charges) above a flat conducting surface.  I don't expect you to rework the problem, you should have the relation handy just in case you need it.  Here it is:


The E field will point downward if Q is positive.

Last week we were looking at how field change measurements at multiple locations could be use to determine both the amount and location of a spherical volume of negative charge neutralized during a cloud-to-ground lightning flash.

The authors of one study noticed that instead of getting larger as you got closer to a thunderstorm the field changes got smaller or even reversed polarity.  This led them to suspect that negative charge plus a small volume of positive charge was being neutralized during the lightning flash. 

Here is what I think would be a reasonable exam problem, one that would make use of the charge above a conducting plane relation above.

Example Question #1
(a)  Compute the electric field change that would be produced directly under a thunderstorm when a -30 C spherical volume
      of charge located at 6 km altitude is neutralized.

(b)  What would the field change be if, in addition to the negative charge, +5 C of charge located at 3 km is neutralized.

Note this problem asks for the E field change, ΔE, not E.  We can make a small change to the earlier equation:

You need to be careful about the polarity of ΔQ. 
 

the left most cloud in the sketch above shows the starting conditions in the cloud before any charge is neutralized.  Then, in the middle picture, we add ΔQs of +30 C and -5 C.  When these are added to the original charge we end up in the last figure with 30 C of negative charge and 5 C of positive charge having been removed or neutralized.

The field change in Part (a) is (you can assume that all 30 C of charge is concentrated at a point 6 km above the ground)

15 kV/m is a large field change, like you would expect to find underneath a thunderstorm.

Next we will superimposed the field change produced when 5 C of positive charge is also neutralized

The field change is smaller.

Example Question #2

Next we looked at Question 2(a) on the Spring 2011 Midterm Exam.  Note this was one of the homework questions in 2015.  Basically the situation is as shown below:



The field at the ground is pointing upward and has an amplitude of 8 kV/m.  The field at 200 m altitude is more intense, 25 kV/m.  You are supposed to find the volume space charge density in the 0 - 200 m layer and its polarity. 

Note in the right figure, you should be able to determine the polarity of the charge from just a sketch.  The negative charge in the thunderstorm will induce positive charge in the ground.  Fields at the tips of pointed objects on the ground may be high enough to go into corona discharge (fields are high enough to ionize the air).  This corona discharge will spray positive charge into the air (actually electrons are "off" air molecules and travel to ground leaving positively charged air molecules in the air).

This is a perfect place to apply the differential form of Gauss' Law.  Here's the solution to the question.

Here's the remainder of the first page of the topics summary


Have a quick look at the calculation we did for a conducting sphere in a uniform electric field (solving Laplace's equation in spherical coordinates).  I would concentrate on the results of the calculation.

I have been looking for some actual conductivity meter/ion counter data so that I could ask you a question about that or perhaps to interpret a graph like shown at the top of the back page of the topics summary.  So far I've been unsuccessful.



Short answer question example #2
We had a look at Question 1(a) on the Spring 2011 Midterm Exam.  The question asks what makes the atmosphere a weak conductor of electricity.  Do electrical currents travel through air in the same way they travel through a wire?


A picture and some commentary are fine for an answer as far as I'm concerned.  Small ions of both polarities conduct electricity in the air.  In a wire free electrons alone conduct electricity.  Ionizing radiation of some kind (you could add additional details here) ionizing molecular nitrogen and oxygen leaving positively charged nitrogen and oxygen molecules.  The free electrons quickly attach to unionized oxygen molecules (but not to nitrogen).  Then water vapor molecular cluster around the ionized nitrogen and oxygen.  The small ions will drift in an electric field at a speed equal to their electrical mobility times E (you could add some numbers).

You could write down the expression for conductivity
 λ = q+ n+Be+  + q- n-Be-

The last part of the question asked why conductivity increases with increasing altitude.  I forgot to try to answer this in class.  And actually I'm not sure I entirely know the answer.  To try to answer the question I would make use of the Table of Electrical Parameters that was handed out in class (reproduced below)

Ion pair production (q) reaches a peak at around 12 to 16 km.  It starts to decrease above that I think because the air density drops.  The recombination coefficient is fairly constant up to 12 km then starts to decrease.  I'm not entirely sure why that happens.  The concentration of small ions (n) stays fairly constant between 10 and 28 km.  The electrical mobility increases steadily with increasing altitude.  Since n stays constant (at least up to 30 km) and Be increases, conductivity should also increase and the table shows that it does.  Why does Be increase?  I'm guessing it is because as air becomes less and less dense, small ions are able to drift further before colliding with an air molecule and slowing down.  The mean free path between collisions increases.

Here's the last portion of the Topics Summary

Note Maxwell current was penciled in at the bottom.  That was the subject of last Tuesday's lecture.

Example Question #3
Here's a question involving the small ions balance equation above.

Given
   an small ion recombination coefficient, α = 1.4 x 10-6 cm3/sec
   a small ion to particle attachement coefficient,  β = 2 x 10-6 cm3/sec
   a particle concentration, Z = 1000 cm-3
   and an ion pair production rate, q = 10 ip/(cm3 sec)
calculate the steady state small ion concentration

We start by writing the balance equation.


At steady state, dn/dt = 0 and we are left with a quadratic equation which we can solve for n.  Substitution in the values of the various coefficients we get

I always get a little nervous about ignoring one root and using the other, but the final answer seems very reasonable.

Example Question #4

This question makes use of your solution to Question #1 on Homework #2.  In that question you were given the fair weather electric field at the ground, 200 V/m pointing downward and the conductivity at the ground, 3 x 10-14 mhos/m.  You determined the potential at an altitude of 20 km, V = +1.6 x 106 volts (relative to the ground).

You are supposed to calculate the capacitance, C, and the resistance, R,  between the ground and 20 km altitude.  Then calculate the decay time constant, RC.

Since we know E at the ground, we can calculate the surface charge density, σ.  We'll then multiply that by the area of the surface of the earth to determine the total charge on the earth.

Next we make use of the relation Q = C V to determine the capacitance (be careful, C in the problem above represented Coulombs)

To determine the resistance we will first multiply conductivity times the electric field to determine the current density J.  Then we'll multiply by the area of the earth to determine the total current flowing to the ground.


In the last step we'll multiply R x C to get the decay time constant.

This agrees pretty well with estimates we've made on a couple of occasions during the semester.