Gauss' Law and Electric Flux

Day 9:  Dipole moments and Electric Flux   

Here we examine the motion of a charged particle in a uniform electric field. We found Fe= qE = ma. When a particle of charge q and mass m is placed in an electric field E, the electric force exerted on the charge is qE. If that is the only force exerted on the particle, it must be the net force, and it causes the particle to accelerate according to the particle under a net force model. 

 The electric dipole moment is defined as the product of the magnitude of charge and the distance of separation between the charges. Here we derived toque(above) which we used to integrate with respect to theta to find work (shown below)

Using the expression for torque which was t=pEsintheta, we plug this into the definition dW=torque*dtheta and integrate it between an angle of θi and θf to find an expression for the work done in rotating the dipole.

here we derived potential energy using a change in energy due to work energy theorem and compared it to gravitational potential energy. 

Here we were given an electric field and we were asked to use Python to predict how the field would look.

Here we have our computer generated electric field. We found that the arrows don't point straight outward but either in or out or side ways, this is because their is a positive and negative charge.

This is where we began our discussion about flux. 

We found that flux is electric field multiplied by area costheta and its units are newtons per coulomb times meters squared.

flux is proportional to charge enclosed over epsilon.

Here we drew an example of how flux would look if it could be seen. Flux is zero when it is parallel to the electric field, and flux is finite when is perpendicular to the electric field. 

Electric field

Here are some ideas we had before we began our first experiment today, which involved using Vpyhon. We also calculated vector r, which we need for our experiment, in the picture above.

Today we started with our study of electric field. We were given an equation: vector E is equal to the F vector divided by the charge on a small test particle. We began the day with an experiment on Vpthon. Given code in which we copied and pasted, we were able to add more code using our new electric field formula and figuring out which way the electric field was pointing from the proton.We found that is was pointing away for the proton. We were also introduced to the "mag" function which is a short hand notation of the calculated magnitude of vector r.
This is our equation for calculating electric field, which is related to electric force (Coulumb's Law).

Here we calculated the net electric field of given points in a system.

Here we calculated the electric field of a point at (0, 2e-10) using Coulomb's Law. Electric field is a vector function so we then found ihat and jhat for the total electric field acting on the pt (0, 2e-10). 

           Here we calculated the equation for electric field due to a differential charge.     

Here we have our work for the next experiment we did, which was finding the electric charge.  we calculated the electric field  corresponding to a continuous charge distribution on a rod at two points in space, P and P'using a spread sheet in excel.

Intro to 3D computer modeling (Vpthon tutorial)

In this lab I was introduced to the program Vpthon. Here I created a model similar to the one in the online tutorial.

This is the code I wrote for the picture above.

Here we have the second computer model I was asked to do with assigned variables.

Made the arrow bigger.

This program was really easy to work with. We just identify our object, for example: sphere, then its position, radius and color. By assigning variables we avoid having to rewrite a long code.

After typing in print and my code this was my answer.

Electric charge

We begin today with the study of electric charge and the forces related to it.

First observation of the day, we have an aluminum sphere which is given a charge. This charge repels the piece of paper stuck to it because of the forces caused by the charge. These forces come from the charge of atoms. The forces between particles that are not moving or that are moving (slowly) are known as electrostatic forces.

Our first experiment:

Electrostatic Forces

Exploring the Nature of Electrical Interactions

You can investigate some properties of electrical interactions with the following equipment.  Each student should have:

• 4 Scotch tapes, approx. 10 cm long

• 2 small rod stands

• 2 metal rods

• 2 right angle clamps

The nature of electrical interactions is not obvious without careful experimentation and reasoning.  We will first state two hypotheses about electrical interactions.  We will then observe some electrical interactions and determine whether our observations are consistent with these hypotheses.

Hypothesis One:  The interaction between objects that have been rubbed is due to a property of matter that we will call charge.   There are two types of electrical charge that we will call, for the sake of convenience, positive charge and negative charge.

Hypothesis Two:  Charge moves readily on certain materials, known as conductors, and not on others, known as insulators.  In general, metals are good conductors, while glass, rubber, and plastic tend to be insulators.

a. You and your partner should each place a 10 cm or so strip of Scotch tape on the lab table with the sticky side down.  The end of each tape should be curled over to make a non-stick handle.  Peel your tape off the table and bring the non-sticky side of the tape toward your partner’s strip.  What happens?  How does the distance between the tapes affect the interaction between them?

b. Place two strips of tape on the table sticky side down and label them “B” for bottom.  Press another strip of tape on top of each of the B pieces;  label these strips “T” for top.  Pull each pair of strips off the table.  Then pull the top and bottom strips apart.

1. Describe the interaction between two top strips when they are brought toward one another

2. Describe the interaction between two bottom strips.

3. Describe the interaction between a top and a bottom strip.

c. Are your observations of the tape strip interactions consistent with the hypothesis that there are two types of charge?  Please explain your answer carefully, in complete sentences, and cite the outcomes of all your observations.

Answer for A was that they were attracted towards my lab partners hands. This is because her hands have the opposite charge of the tape.

answer for b1 is that they repel each other, this is because they have the same charge. 

answers for b2 and b3: b2: repelling; b3:slightly attract.

summary of our experiment. answer for C is yes, attract and repel.

This is our experiment in which we review a little of physics 4A. A ball hangs vertically by a string of length L. Initially ball is located at the origin of your coordinate system. The ball is pushed horizontally so that now it is in a new position with x-coordinate x2. Write an expression for the angle θ makes with the vertical.

The magnitude of the gravitational force between two masses m1 and m2 

Electric Force Law Video Analysis Activity 

Here we used logger pro to measure the electric force between to metal balls, one hanging and one on a stick. We got the one on the stick closer and closer to the hanging one. The hanging ball started to move away as because of this. From there we were then able to find the separation distance caused by the electric force.

b)Determine the percent difference between your experimental value for the exponent in this force law and the theoretical value.

Assuming that the charges on both balls are the same, use one of your data points and Coulomb’s Law to determine what the charge is on each ball.

Assuming that the charge on the hanging ball is half the charge of the other ball, use one of your data points and Coulomb’s Law to determine what the charge is on each ball.

Here we calculated the the magnitude of the force between the two charges.

Here we calculated the angle in which the string moved



Here a propeller was placed on top of the charged sphere once placed on top the it started to propel and the pieces of paper on the sides became neutral.

Here we calculated the the ratio of the electric force to the gravitational force between two particles. We were given a pair of electrons with a charge of 1.602x10^-19 C and a mass of  9.1*10^-31 kg  separated by a distance of 1 * 10^-10 m.

Heat Engines and Entropy

here we have one of our engine simulators, in which we place the model on top of a breaker which is cold on top and heat coming from the bottom. The transfer of heat between the two causes the "engine" to propel. This is caused by entropy, the temperature does not raise, but there is disorder.

These are the derivations we came up with for our calculations. we found Q_C, Q_H, and work, then combined them to create an equation we needed to find the final temperature of the system.

here we asked to find the final temperature of the experiment, which we found to be 46 degrees celsius.   the Carnot engine is the most efficient engine to date, however, it can not be used in the "real world". the Equation in the picture above depicts this. here we calculated the efficiency of our engine in the first experiment.

Here we discussed the diesel engine and its cycle. In a diesel engine the pressure compresses more and thus extends adiabatic process, as shown in the shade region of the graph. When compared to a gas engine, a gas engine runs faster than a diesel engine but it is more efficient.

DIESEL CYCLE

n moles of an ideal Diatomic gas undergoes the following 4-step cycle:

1 to 2: Adiabatic  compression

2 to 3: at Constant pressure

3 to 4: Adiabatic expansion

4 to 1: constant volume.

For each “corner” of the cycle calculate P, V, T.

For each process (A to B, etc.) and for calculate Q, W, ∆U, and ∆S.

Calculate total Q, W, ∆U, and ∆S for the entire cycle.

Calculate the efficiency of the heat engine running this particular cycle.

Given: p0 = 105 N/m2

V0 = 10-3 m3

T0 = 300 K

Qin = _600_ J

V2 = 1/_15_ V0

CV = 5/2 R; CP = 7/2 R; γ = 1.4

These are some of the calculations we did for our diesel engine, which also includes the coefficient of performance.

here is our calculation for how long it would take for water to freeze, given the mass of the water and the heat energy.