Thursday Jan. 22, 2009

We'll spend at least a couple of weeks reviewing some basic laws from electrostatics.  Examples from atmospheric electricity that make use of some of these principles will be inserted along the way.

We'll start with Coulomb's Law.  Many of the figures below were redrawn after class for improved clarity.


The force that q1exerts on q2 is given below and depends on the product of q1 and q2 and varies as 1/(distance)2


The principle of superposition applies: when multiple charges are present, the force exerted on one of the charges is the vector sum of the forces exerted on the charge by all the other charges.  Note the vector and unit vector notation being used above.  We will mostly be using the MKS system of units in this class.

If a charge q is placed in the vicinity of a charge Q, we could use Coulomb's law to determine the force that Q exerts on q.  We can imagine a vector field, the electric field, existing around Q even before q is brought into the picture.  Multiplying q times E would give the force that Q exerts on q.  The expression for electric field is shown above.


Because the curl of the vector r divided by r3 is zero, we can replace that term in the expression for electric field with the gradient of 1/r.

This leads to a definition of the electrostatic potential.  It is often much simpler to determine the electrostatic potential because it is a scalar quantity.  The electric field can then be determined by taking the gradient of the potential. 

The expression above is valid for a point charge.  More general expressions for cases where multiple charges are present or when charge is distributed over a volume or on a surface are shown below.  This figure was not shown in class.


Essentially you would need to sum over a collection of multiple charges, or integrate over volume and surface distributions of charge to determine the electrostatic potential.


We'll try an example problem: calculating the field a distance r away from an infinitely long line of charge.

Short segments above and below z=0 contribute to the field.  The z components cancel, the r components add.

We'll come back to this problem and solve it in a much simpler way.




This is the integral form of Gauss' Law.  We'll return to our earlier problem and uses Gauss' Law to determine the electric field.  We'll see that it is a much easier process.

We draw a cylinder around the line of charge.  This is the area that we will integrate E over in the Gauss Law expression.

There is no contribution to the integral from the ends of the cylinder (E is perpendicular to the normal vector, the dot product is zero).  E is parallel to the normal vector along the side of the cylinder.  E is also constant on the side of the cyliner (E is a function of r and r stays constant as you integrate over the surface of the cylinder side).  In the end we obtain the same expression for E as we did in the earlier example.
Here's a little more detailed explanation of two of the demonstrations conducted in class last Tuesday.

The Volta Hailstorm apparatus consisted of two metal disks mounted above each other.  They were mounted inside a clear plastic cylinder.   Small round balls made of aluminum foil were inside.  The apparatus was placed on top of a Van de Graaff generator.   The foil balls are in contact with the top electrode and acquire some charge (assumed to be positive). 


The balls are repelled by the bottom electrode and travel up to the top metal disk.  They transfer their charge to the top disk and then fall back to the bottom disk.

The charge on the top disk bleeds off into the air through the pieces of wire connected to the disk (the charge would eventually travel to ground and complete the circuit).  The foil balls again acquire charge from the bottom disk and the whole process repeats itself. 


The Franklin chimes demonstration consisted of a center bell that was connected to the Van De Graaff generator and bells to the right and left that were grounded (only the left bell is shown in the figure above).  A small metal ball hung from an insulating thread between the center bell and the left bell (and also between the center and right bells).  The small ball is assumed initially to be uncharged.  How ever the charged center bell will cause charges to be induced on the surface of the ball as shown in the figure (these charges make the electric field inside the ball zero).  The ball will be drawn toward the center bell.


Once the ball contacts the center bell, the negative charge is neutralized.  The positively charged ball is repelled by the center bell.


Once the ball touches the grounded bell, the positive charge flows to ground. 


The ball is again uncharged.  The whole process repeats itself.  The motion of the ball is transporting charge from the center bell (i.e. from the Van de Graaff generator ) to ground.