Extra Credit for Test (Raman)

I spent my Friday afternoon learning about Raman spectroscopy. Proof that I was there. Photo bombed by Mike E. It was very fun and informative. Too bad it is not part of my major.

Day 11 Gauss Law

Gauss Law = q,enclosed/eplsillon not. We divided a couple of charges into different enclosed areas, s1, s2, s3 and drew the flux lines from each charge.

We had q1 be negative( top right), q2 and q3 are the positives one. Flux lines are going towards q1. The flux for each S (surface), s1 ,s2 ,s3 can be counted by having flux lines going out - flux lines going in. If the net result is positive then more flux went in the surface, if net result is negative more flux went out of the surface.

Another picture demonstrating flux.

Electric field only effects the normal of a surface.

Flux inside is zero in this enclosed surface.

Prof. Mason then asked us what would be the best answer if one was in a lighting storm. Our answer was to stay inside the car. It would act like a Faraday cage. 

The white board Prof. Mason picked out.

We then explained the demonstration of a lighter in the microwave. And when to use the right symbol for E calculations.

We explained the plasma and also E = sigma/Epsilon not. And when to use lambda, sigma or rho in our E calculations. We also explained what the fork did in the microwave.

We then derived the volume of a sphere and integrated in terms of  r + dr.

The white board Prof. Mason showed as example.

Our work of the same thing. dr to the powers of 2 and 3 are small so they go to zero. Only 4pir^2dr is left and is useful.

We then did some Active Physics questions that asked to find Q if the r's are different. The final answer was 1/8Q.

We set the lesser r on top and the bigger r on bottom. Did some canceling and was left with 1/8Q.

We then did use Gauss law to find E.

E = q/e0, the 4pi and e0 turns into K. E then becomes E= Kq/r^2. Similar to Coulomb's law.

Prof. Mason then pleased us by showing us what different types of things do when they are put inside a microwave. We did some prediction for some.

Fork, our prediction was the fork would cause sparks. Nothing happened.

Cd, our prediction we would see sparks. Sparks happened.

The damage to the Cd by the sparks caused by the small layer of aluminum coating on it.

Our prediction of Soap. Did  not work, but the soap would become foam.

Picture of something.

Picture of soap.

Very beautiful colors when microwaved. Melted a whole through it.

Picture.

Our predticion was that it would expand and blow up. It did not but showed a light blue glow.

We then moved on to answer some Active Physics demo and questions. The first was a slew of questions about flux.

All questions answered. Used Gauss law to find flux. Which was 40,00N/C. Flux in a smaller radius is zero. The charge is 2.39*10^4 C/m^3. Rho*V is equal to E is = 45,000 N/C.

Prof. Mason then asked us to E for a cylinder. Sa = 2pirL + 2pir^2. Lambda  = q/l.

We plugged in the numbers to the equation and got E = 4.798 N/C for the cylinder. Used q/L. The second term goes to zero since the is no perpendicular flux going through it.

For got to mention the graph we found from E(r). E field grows as the radius grows.

The E is greates when little r is equal to R. After little r becomes greater then R, E drops by 1/r^2. E(r) = qnotr/R^3.

Summary:
We found Gauss law to be useful in finding E. Flux inside different S have different flux depending on the q inside and the q outside of the S. Flux is = integral EdA. The usefulness of 4pir^2dr found for spheres. Changes in r could effect Q greater then one would think. Gauss laws is similar to Coulomb's law.
Microwaved a whole bunch of things.
Learned that little r inside of big R stores different Flux. Found the graph of E versus R.
Found the Surface area for a cylinder  Sa = 2pirL + 2pir^2. Lambda  = q/L.

Day 10 Electric Flux

We had to draw a picture that represents electrical flux and what it meant.

Flux can be though of the lines of electricy passing through. If lines going in are = to lines going out flux is zero. If there is any disrepcancy positive or negative then the flux if positive or negative.

Dipole if q+ and q-. Finding if there is going to be a torque.

We did some math to find an expression for torque.

Continuation of work, we used cross product r x f.

Picture of Prof. Mason work of dipole torque. Much cleaner and easier to see.

Torque becomes - 2aqEsintheta

We then did some Active Physics lab and questions. One of the demo question was to find work as it spins.

Work is equal to rho*E(costheta inital - costheta final).

We then had to find Ey and find an expression for it.

Ey = KQD/ (d^2+ a^2)^(3/2). This expression was similar to the one of gravity from 4A.

 We then moved on to a demonstration of the Van de Graaff generator. We had to make some prediction of what would happen to aluminum out side. We said the ones inside would stick inside and the ones out side would repel against the cage.

Picture before its on the device.

Shows that the ones inside are stuck to the wall of it and the ones outside are repelled by it. Our prediction were right.

Calculus work from before and our prediction on white board.

We then moved on to working out some Active Physics problems. We had to draw flux lines when there was different q.

This one had q, 2q and 4q thus the there had to be more lines extending outwards.

We then had to draw the connection of different charges and their flux.

We had +q and -q. The flux lines went from positive to negative.

Prof. Mason then showed how cool he was by setting on spikes. They represent flux.

Where he is going to sit.

Him being a boss.

We then had to answer some question from the demos by Active Physics lab. The following white board have the answers.

Playing with the Active Physics demo; Answer to number 1.

Playing with Active physics demo; Answer to number 2.

Playing with Active Physics demo; Answer to number 3.

The same as above, Answers for 5,6,7 and 8.

Flux is equal to EA = E4pir^2
We then found the flux of a cube.

Only the E on the left side and right side matter. All the other sides had no E "lines" going through them thus they were zero. Flux then became EA - EA which turns out to be zero.

Summary:
A dipole experinces a toruqe. Torque is = -2aqEsintheta. Work is = rho*E(costheta inital - costheta final).
We found an expression for Ey =  Ey = KQD/ (d^2+ a^2)^(3/2) which looks similar to the gravity integral we used in 4A.
Did an experiment with the Van de Graaff generator. Our predictions work correct. Inside sticks, outside repels.
Did a whole bunch of Active Physics Demos. Learned that Flux is zero when there is no charge. Flux is related to the magnitude of charge. Flux depends on the summation of charges inside a ring. Flux = EA.
Flux only effects areas that are on its "way".

Electric Field Hockey Assignment

Electric Field Hockey. Complete level 1, 2, 3. And

Level one easy.

Level two took a little longer.

post pictures of completion. Level three is impossible!

Day 9 Electric Fields

Electric fields are electric force per charge. We typically use positive test point charge thus electric fields radiate outward.

We were asked to explain gravitational force of attraction and compare to electrical fields. As a group it was hard to explain. The field of G and the field of E are similar. We came up with the conclusion that electric fields are caused by charges, the magnitude of the electric field depends on the magnitude of the charge and magnitude of the electric fields depends on distance as well, lastly mass creates charge.

We then had Active Physics lab questions. We had to use E = F/q for each point and each electric field.
White board work is missing. :( But conclusion was the negative number means is inward field and positive number means outward field.

Prof. Mason then demonstrated with charges and electrical fields with how a marvel moves around. We came with the conlusion that the marvel moves in circular paths in and out of the center. Anwser to number 5 is the same or uniform becuase the eletric field is uniform. And anwser 6 is that force changes direction towards +q.Again, the video and pictures of the demonstration is missing. :(

We then did the Active Physics lab.

The final result was that the F = <18.1 N/C, 12.8N/C>

We then did the example of a uniform bar charged at a distance away.

Did some white board work. The answer for E was cut out but was found to be 52,600 N/C.

We then used a spread sheet to find the vertical  component of charge.

Visual diagram of what was happening.

The charge was found to be 1.27*10^5 N/C from the spread sheet.

Picture of spread sheet.

We then had to do some calculus to find an expression for Electric fields.

Found the expression for E. We can plug and chug from this point. And it turns out to be 5.9*10^4 C.

Summary:
Defined what Electrical Fields were and how they relate to Gravity Fields.
Found how E = F/q. If E is negative the field is inward, if E is positive the field is outward.
Did some Active Physics problems to find E and Forces. Used a spread sheet to find q vertical of a uniformed charge rod.
Dervived an expression for E to be used. KQ/d(d+L).

Day 8 Electrostatic Forces

Electrostatic force was demonstrated by Prof.Mason using balloons. He would wipe them with animal cloth and give it a charge. Then asked us if it would stick onto glass. We said it would and it did. Prof. Mason then asked us to to draw a FBD of the balloon stuck on the glass.

We draw the FBD. Friction is up, mg is down. normal is perpendicular to the glass and the Fe is inward to the glass.

We then drew the charge distribution on white board representing the balloon.

Plus charges on the side of the glass, negative on the other side of the balloon.

We then did some tape experiment to find if there were more then two types of charges. The answer is that there is only two types. Positive and Negative. We used tape and scrubbed it in till one was positive and the other was negative.  Similar charges would repel, and different charges would make the type attract. Unfortunately we do not have the video or pictures of this part of the lab. :(

We then had a Active physics lab. We had to find the force the charge would repeal the ball.

We found Tension and the change of theta. F = mgtan(arcsin(dx/L)

We then used Logger pro to analyze the video. We found two fit equation for the data  from Logger Pro.

We fitted the equation to be F =AR^-2. Where A is a constant and R is radius. A = 0.1421.

The second fit equation, A = .1204 B = -1.934.

We then got those numbers and plugged them into Coulombs' Law. We showed that the force was portional to the square  of distance becuase of the way the graph looked. We also found q, and our percent error.

q = 1.56 * 10^-15C. The percent error was 3.3%. And the graph plotted showed a 1/r^2 similarity.

The last part of the lab was how charges interact with one another. We applied Coulombs' Law.

Some work was erased. But we found the F = -13.5*10^-6 N and the r vectors <-4.26*10^-6 N, -12.80*10^-6 N>.

Prof. Mason then did some demonstration with Van de Graaf Generator. To show the power of electrostatice force. Unfornuatley the video and pictures are missing.

Prof. Mason then demonstrated the Lighting Ball sphere. Again the video and pictures are missing.

We did some Active Physics problem to relate electric and charges in different situations.

White board work, we found K to 8.99*10^9. We divided Fg/Fe

We found Fe = 4.2*10^42. Much much much bigger then Fg.

Summary:
We talked about electrostatic force, had an example of the balloons sticking the the glass. Drew a FBD of the balloon.
Found an expression for F for the Active Physics lab. Plotted how the chrage ball would displace the other. Fitted equations that would find the constants A and B. Found that the experiments follows the behavior of 1/r^2. Our error for the experiment was 3.3%. Found the q to be 1.56*10^-15 C. Found the F to be -13.5*10^-6 N. And the r vectors <-4.26*10^-6 N, -12.80*10^-6 N>.
Then Prof. Mason Demonstrated  Van de Graaf Generator and the Lighting Ball Sphere. Pictures and Video missing. :(
We then lastly worked out how Fg compared to Fe. Fe is much much much bigger than Fg.
Fe = 4.2*10^42.