Flux and Gauss Law

Introduction, Today we learned about Gauss Law. We learned that flux = E.A or flux is equal to the closed surface integral of the electric field dotted with dA (the areas that makes up the entire surface)

or the charge enclosed over some constant epsilon not(q/epsilon). 

Note net electric force is always the same anywhere in the field defined so we can relate F=qE=ma

Here we have two charges to which we find the electric dipole moment, which is similar to moment of inertia. So what we did was let F= qEsin(theta) and then say r= d/2. so torque is F(d/2) + F(d/2), or qEdsin(theta) or pxE. potential energy is qrE. (think m=q, h=r, E=g)

In conclusion the Flux is dependent on the surface area or area perpindicular to field and the strength of the field or E.A. The electric dipole moment can be said to equal pxF.

Electric Field

Introduction: In today's lecture we learned about electric fields. We learned about their properties, and how to derive this field E. Where E is Force/Q2 and can be rewritten as E=kQ/r^2.

Answers to the questions from online. Answers to basic questions, to test understanding of the Electric Field and how they react differently to positive and negative charges.

We are asked to find the E-field at these 4 points.

Here we show the total E at each point. E=kq/r^2. We wrote an equation so that wherever you place your two point charges (x,y) and (x1,y1) for any charge q1 and q2 it can find the E-field at any point (Px,Py).

In Conclusion, Every particle with a charge emits an electric field, no matter how far. However it is inversely proportional to distance, so many times the electric field is negligible if the distance is really large and the charge is small. There are two types of charges that play a role in electric fields and that would be a positive charge and a negative charge the negative charge looks like a bunch of vectors pointing towards it, if you were to look at how the electric field around a charge would behave. A positive charge would have vectors pointing radially outward.

Electric Force

Chris Minaya, Michael Pryor, Andrew. Pete

Introduction: In this lab, we will observe the force between two charged objects. We will examine a video where a ball hanging from a string is slowly subjected to another ball that has the same charge.then the ball hanging from the string is repelled.

Here we derive an equation for the Force acting on the ball hanging from the string.

Conclusion:
1.) Yes, we were able to tell immediately that it was an inverse relationship. Then by looking at the slope we figured it must be and inverse square due to the steepness of the graph.
2a.)  They are the same value, so the percent difference is 0. However our equation for the function does not accurately describe the function too 100%.
2b.) A=kq^2 => q = sqrt(A/k)= 5.25 * 10^-8
2c.)A=k(q/2)(q)=k(q^2/2) => q = sqrt(2A/k)= 7.43 * 10^-8 and q/2 = 3.72*10^-8
3.) No, we derived the force in the x direction from using a fbd and equations for static equillibrium. Since this is what we used for our force equation our force will only be positive since distance is positive (measure from origin). Hence the best we can do is find the absolute value of q.
4.)Uncertainty for force can be found, systematic error can be found for not accounting for the mass of the string string and small things like friction and minute spinning of the string. We also assume that acceleration is equal to zero.

Stirling Engine

The hot reservoir is located on the bottom side of the apparatus and the cold reservoir is on the top side.What we predicted would happen when the ice was placed was that the engine would take energy from the hot reservoir Qh and use it to do work(W), while Qc would be exhausted to the ice. The larger the disparity in temperature the more efficient the engine becomes. What actually happened is exactly what we predicted!

Carnot Cycle & Engines!

Here we are shown a PV graph and asked to find the W.

Close up shot of Values

Unfortunately we were not able to solve for the W in class, however we did learn that work is the area in the shape. So we should have approached it by seperating the Graph into Trapezoids i would use 6 one from 32-35 the other from 35-37 and lastly 37-40, then i would subtract the bottom trapezoids giving us 6 in total.

Here we can use our definitions of internal energy work and heat, to solve for the constant Cp=5/2R.
We can also do the same for Cv by letting work = 0 and Q=nCvdelT giving us Cv=3/2R
Note: Cp-Cv=R

Here we have an adiabatic Expansion with small changes in pressure and volume. we use the chain rule to solve for ndelT, and then we substitute our Cp-Cv for R.using our Definition for adiabatic expansion we let Q=0 then set our definitions of Energy Internal to write a new equation. From here we set our two new equations equal to each other and take the the integral.

Now we can show relationships between P T and V. by using our ideal gas law PV=nRT.
Note: gamma = Cp/Cv

Using our new Formula for Pressure initial we can now solve for W=Spdv.

Here we solve for DelE Q and W by using our new formulas we just derived for the adiabatic expansion part of the carnot cycle. There were a few givens like Q=0 for adibatic and for Isothermal (U=3/2nKbdelT) Internal Energy is 0. From there we just used Wadiabatic to solve for W and E internal for two of the processes. The Isothermal Processes we used Eint = nCvdelT and then solved for W since Q=0. To find the efficiency of the system  we can W/Qh.

Quiz

Givens:

Work for P,V and T for points 1-4

Here i did it with n moles = 1, i did it another way as well but the number are ridiculous i will post it beneath

Givens and Work for Q,W and ΔU for all processes.

w=-80J and delU = 80J

In Conclusion, del U is not zero. but monday is over! 

Work for Thermal Efficiency.

Heat Engine and Cycles

A candle burning the oxygen and suffocating itself.

A candle burning the oxygen that it gets by induction which keeps the flame burning.

Top: Graph of an Isobaric Process where V and T are the only variables, notice how their is a linear correlation.
Bottom Left: Graph of an Isochoric Process where p and T are the only variables, and have a linear relationship.
Bottom Right: Graph of an Isothermal process where p and V are the only variables, and have an inverse relationship.

Now that we understand the relationship of the Isobaric, Isochoric and Isothermal Process, we can set up an equations where we have one pair of variables at time intial equal to those a t-final. By doing this we can just solve for whatever variable we did not measure or have originally!

Notice how we can find the pressure at any given volume, since we were given pressure and volume initial and were told n and T were constant. here we first solved the pressure at 20dm^3 then at 10dm^3.

PV graphs of isobaric, Isocholoric, Adiabaitc, and Isothermal Processes.

We solve for the work of gas by taking the integral of Pdv which gave us P_V_ln(V/V_).

Next we would solve for the W done by gravity which is (3/4)mgh for this specific case since the water tank was only 3/4 full of water. This also means that there is air in the tank, which means their is work being done by the air. The work air exerts is (1/4)*atmospheric pressure * 50. Once we add all of these values we finally get our answer or 1.793 *10^7 MJ.

Here we have a problem where Volume and Pressure are variables. So what we did was take the integral of P(v)dv but substitued nRT/V for P(v) which we can use an integral to describe its change since the only thing changing would be V. From there we can substitute PV back in or leave it as nRT. 

A PV graph was given to us which contained 4 points, the first thing we did was observe when work was being done by the gas(when P or V increase) and when work was being done on the gas(when P or V decrease). Then we went ahead and found the internal energy at each point which we describe to be equal to (3/2)PV.

From there what would be done is to label each of the sides A,B,C,D and let each side represent the change in internal energy which we could solve for by subtracting the final point from the initial point(like vectors). Finally we could find the Work done by using pdelV, then use the first law of thermodynamics to solve for Heat(Q). We could also solve for Worknet by adding the work from each side.

Here we have the efficiency of our system, which can be solved for by taking the ratio of Work over Heat added to our system.