Physics4B vlgomez: March 31 Gauss' Law and Electric Field

We began class with the activity "Gauss' Law in Flatland." We opened Coulomb program online that helped us create an electric field, we then sketched "E-field" lines with 3 charges of low magnitude. The picture above is our hand drawn interpretation of what designed on the computer. We then drew arrows on the electric field lines to indicate the direction in which a small positive test charge would move. We then drew different 2D "closed surfaces" around the charges.

We continued our experiment by counting the net flux lines coming out of each "surface." Flux lines going into a surface are negative and flux lines coming out of a surface are positive. The net number of lines is defined as the number of positive lines minus the number of negative lines. The picture above shows theses calculations.

We found that the relationship between flux and net charge is that they are proportional to the charge enclosed. In the picture above Gauss' Law is defined.

Here we began our next experiment that involved a Faraday cage. There are strings attached along the top of the cylinder. Inside and outside sides of the cage are rectangular aluminum foil chunks suspended from those strings. The cage is then connected to the van de Graff generator. The generator will put in a large negative charge onto the cylindrical screen. The question we were asked was, what happens to the foils when the generator is turned on. My groups prediction was letter C, both the inner and the outer foils will move away from the cylinder.

The result from the experiment was that only the outside foil is moving away from the cage and the foil inside the cage did not move. The foil inside did not move because the electric field inside is zero. This is because the charges only gather at the furtherest distance possible, which in this case is the outer part the the cylinder.

Here we depict how the charges act on the cylinder and thus explaining why only the outer foil pieces moved away.

In the next part of class we were given a hypothetical scenario.

You are trapped in a lightning storm in your car. What is your best course of action. Give a detailed description to support your answer. 1)Get out of the car and run to the nearest tree2)Get out of the car and lie down flat on the ground.3)Seek the highest point nearby and put up your umbrella4)Seek the lowest point nearby preferably a ditch or ravine.

5)Stay in your car

The answer was 5, stay in the car. This is because the charges are only along the outside of the car and there is no charge inside of the car, thus keeping the person in the lighting storm safe and alive.

An electrical conductor is a material that has electrical charges in it that are free to move. If a charge in a conductor experiences an electric field, it will move under the influence of that field since it is not bound (as it would be in an insulator). Thus, we can conclude that if there are no moving charges inside a conductor, the electric field in the conductor must be zero.

Here we did a little geometric review of circles and spheres. In the picture above, we the work we did in class for the questions listed below.

Activity: Some Geometry of Circles and Spheres

a. What is the equation for the circumference of a circle of radius r? If the radius doubles, what happens to the circumference?

b. What is the equation for the area of a circle of radius r? If the radius doubles, what happens to the area?

d. Find the derivative of the volume V as a function of r. Show how this derivative can be used to determine how much the volume of a sphere would increase (that is, the factor dV) if the radius of the sphere were increased from r to r + dr. Hint: Consider this increase as being the volume of a shell of thickness dr surrounding the sphere.

e. If the letter S is used to represent the surface area of a sphere, what is the volume dV of a thin shell of thickness dr that surrounds a sphere of radius r in terms of S and dr?

f. Use the derivative dV/dr from part d. and the idea that a spherical shell represents a volume increase to show that the surface area of a sphere can be represented by the equation S = 4πr².

g. If the radius of a sphere doubles, how much does its surface area increase?

h. If a charge Q is spread uniformly throughout the volume of an insulating sphere (that is, charge cannot move around inside it) of radius r, what fraction of the charge lies within a radius of r/2? Warning: The answer is not 1/2.

Here we answered the question, "what is the magnitude of the electric field, E, as a function of the central charge,q, and the distance from it, r?

In the picture below, we used Gauss's law along with the equation above to show that the magnitude of the electric field inside a uniformly charged sphere of radius R having a total charge of Q is

In the next part of class, Professor Mason took out a microwave. He then microwaved a fork, steel wool, and a CD. In each demonstration we seen sparks come out of the tips of the fork and steel wool. Sparks also came out of the CD. This experiment proved that electric charges do gather at the outer part of the metal conductors. The pictures below are from the experiment.

Next activity we did was from our lab manual and it asked the following questions:

Activity: Some Geometry of Cylinders

a. Consider a cylinder of radius r and length L. What is its volume in terms of π, r, and L?

b. If a charge Q is spread uniformly throughout the volume of a cylinder of radius r and length L, what fraction of the charge lies within a radius of r/2? Warning: The answer is not 1/2.

c. What is the surface area of the cylinder in terms of π, r, and L? Hint: Don’t neglect the ends.

Our answers for the questions above are pictured below:

Next activity we did was Gauss' law and Cylindrical Symmetry

Activity: Gauss’ Law and Cylindrical Symmetry a. Use Gauss’ law to calculate the electric field at a distance r from a very long, straight, uniformly charged cylinder that has a charge per unit length of λ. (Mathematically, we can treat a wire which is physically very long as if it were infinitely long.) Hint: Using a symmetry argument, explain why you can neglect the electric field perpendicular to the two ends of the cylinder.here we calculated the inside of the cylinder.

Here we calculated the outside of the cylinder.

Last activity we did for the day was The Gravitational Gauss' law. Which shows us the relationship between electric field and gravitational field. The fact that electric field lines spread out so that their density (and hence the strength of the electric field) decreases at the same rate that the area of an enclosing surface increases can ultimately be derived from the 1/r² dependence of electrical force on distance. Thus, Gauss’ law should also apply to gravitational forces.