The Midterm Exam Review handout was essentially a "mini" midterm exam.  We were able to work through most of the question during class on Monday.

The format of the "real" Midterm on Wednesday will be similar.  Question #1 will consist of several short answer questions.  You will be asked to choose and answer 3 questions and will be able to pick from a total of probably 6 questions.  You should plan to devote 5 minutes or less to each of these questions.  They will all have the same points totals, so it makes sense to answer the easiest questions or questions about material you are most comfortable with.

You will then be asked to answer 3 longer questions that might require a derivation, a calculation of some kind, or a more detailed explanation of a phenomenon or a concept.  You should generally plan to spend 10 to 15 minutes on these questions.  And again you will be able to choose from a total of 5 or 6 questions.

The Midterm Exam is Open Notes.  Make sure you are familiar with what is in your notes.  You are free to access the Online Class Notes that accompany this course.  Many of the questions were designed with that in mind.  You do not need to redo material that is in the notes or that you have worked out already on a homework assignment.

Here, as best I can remember are the answers to the short answer questions.

#1e.  This was, in my mind, the easiest and shortest of the questions.  A "Leyden Jar" is basically just a capacitor.  A sketch is included below:

#1 d.  A good place to start is to write down the small ion balance equation and identify the production and loss terms:

Then set the recombination and attachment loss terms equal to each other, plug in values for n, α and β .  The resulting value for Z, the concentration of particles seems very reasonable.

#1b.  This is a very quick question to answer if you know where to begin.  We'll use the continuity equation:


We assume steady state conditions and the J has only a vertical (z) component.

1a.    This was one of the longer answers in this first question.  Basically a plate of metal with area A is exposed to an E field.  The E field induces charge on the surface of the metal plate.



In the case of an electric field mill the sensor plate if periodically covered and uncovered (the area of the plate exposed to the field will change).  Charge will run to and from the plate producing a current.  With a fast or slow E field antenna the plate is always exposed to the field (area stays constant) but the field itself can vary.  This will again cause a current to run to and from the plate.


The signal current is proportional to dA/dt in the case of a field mill and proportional to dE/dt for the fast and slow E field antennas.  In both cases the signal current is integrated to give an output voltage proportional to E and A.

The integration can be done using either a passive or active integrator.  The decay time constant of the passive integrator will be determined by the input impedance of the measuring device the antenna is connected to ( decay time = R C).  In the active integrator the decay can be adjusted using R and C and is independent of the measuring or recording equipment.

1c.  Here's the last part of Question 1.

Let's first look at the polarity of the fields (a).

Positive polarity fields are what you would expect to see under a thunderstorm.  The E field points upward toward the main layer of negative charge in the thunderstorm.

The fields aloft are much stronger than the fields at the ground (pt. B).  This is because the E field at the tips of pointed objects at the ground are enhanced.  There is enough enhancement to break down the air in the vicinity of the point.  The air near the ground is filled with positive polarity space charge.

The positive space charge above the ground reduces the amplitude of the field at the ground.  Note also how the fields recover after the lightning discharge (the field change at pt. C).  The recovery is linear in the fields measured at 200 m, sort of an exponential decay in the fields recorded at the ground. 

Finally there is the field change itself.  It is in the negative direction.  



Most cloud to ground discharges removes negative charge from the cloud and carry it to ground.  The next figure tries to explain why this would produce a negative going field change.


It's as if you added a small volume of positive charge, ΔQ, to the charge that was already there.  The volume of positive charge would produce a negative ΔE.

Notice also that the amplitudes of the field change on both field recordings are equal.  The field change occurs so quickly that space charge can't be added to the air next to the ground quickly enough to mask the field change.

2.    We write down the small ion balance equation again and set the dn/dt term equal to zero (steady state).  We're left with a quadratic equation.  The solution to the quadratic is given in the notes that I put online from the 2013 Midterm Exam Review.  The would be the best approach but might be somewhat time consuming and you would need to be careful not to make any algebraic errors.

Something else occurred to me.  In the case of clean air let's just neglect the particle attachment term.  It's pretty easy to solve for n in that case and you get n = 2673 cm3  .

In the case of dirty air, we'll ignore the recombination term.  You can quickly solve for n and get 1000 cm3



Note that our quick solution isn't to bad.  The solutions for n in clean and dirty air using the quadratic equation are shown at the bottom of the page above.

5.   We really didn't answer this question in class.  Though I did mention there will probably be a qualitative question like this on this semester's midterm.  I think you'll find everything you need to answer a question like this embedded in the Jan. 16 class notes.

3.    You can take care of this question pretty quickly if you know how to proceed.


They key step is using the differential form of Gauss Law to replace the partial derivative of volume space charge density with εodE/dt.
4.    We are going to make use of one of our homework solutions in this problem, the point charge above a conducting plane (which made use of the method of images)


Instead of a point charge Q we have an increment of charge dQ = λ dz  [ λ here is the line charge density (C/m) not conductivity ]



Because we are calculating the field directly below the leader channel, D is zero.  We integrate over the length of the channel to find the total field.



It's a pretty intense field, but the bottom tip of the leader channel is only 100 m from the ground.  I'm not entirely sure the sign is correct.  The E field should point upward toward negative charge in the leader channel.