lab13 April14 (power and potential)

We built closed circuits using light ball and batteries and drew circuits diagram.

Electricity generates energy. We put a resistor in water and measure how the temperature changes with time. We can calculate the heat and energy generated by the resistor usingQ=mc*deltaT

This is a problem given resistor and voltage for calculating current.

Recall Physics4A, how to calculate the work to pull a cart to a certain height using three different paths? 

Finally we find the work are all the same no matter what paths the cart takes.

Compare which of the three paths does the most work? Provided path A,B,C are same length.

We know V=kQ/r. If there are more than one point charges, Vtotal=V1+V2+V3+V4...Vn

For closed circuit, electrical potential total is always zero.

Draw equal potential lines surrounding a center of point charge.

Then, we solve a problem which applies the Vtotal=V1+V2. We solve how more than one point charges affect one reference point and calculate the electrical potential.

Lab 12 April 8(Current voltage and resistance)

Question:

What is the simplest way to describe energy? 

Answer:

It is the capacity to do work.

Question:

Why do you need a wire to go back from the bulb to the battery?

Answer:

The electrons can flow in and out of the battery causing an equilibrium charge.

Question:Suppose there is a hammock maker. What are the 2 things need to know about waterfall?

1. Height, which related to the potential energy/pressure

2. Rate of flow

Similarly, we use hammock maker to understand electricity generator.

Voltage is the height of the electrical cliff.

Height is the voltage(energy/unit charge), flow rate is the current(charge per second),power is the energy per unit time

We are introduced to the P=VI (power=voltage*current)

Then we use this to solve a problem, which is given current and voltage calculate the power of the closed circuit.

Ammeter is a machine used to measure current. It is old-fashioned, but very efficient. 

Our purpose is to measure both currents in wire on top and bottom of the bulb, and compare them.

By "top" and "bottom" we mean that placing the ammeter firstly between the +pole of the bulb, and secondly between light bulb and the -pole of the battery

Over a wide range of conditions, the flow of the charges quickly achieves a steady state value and remains constant. The "average speed" at which the "free" charges are moving in the wire is called the drift velocity vd.

Given area,I ,density . We want to know the drift speed.

Next, we are going to measure resistor using Ammeter and Volt meter.

Graph the relationship on logger pro, we find some of the resistors follow Ohm's law, while some are not. Some of the graph are linear relationship, some are not. Why?

Prof Mason says it is because 1,the light ball turns on 2, material change, (non-Ohmic material)3. material dependent as voltage increases 4. temperature. etc. 

**What does resistance depend on**

Lenghth :160cm and 200cm nickle. How should the R compare? The longer the material, the bigger the resistance. We conclude that resistance is proportional to length. We use similar methods to find that resistance is inversely proportional to the cross section area of the wire.

Temperature increasing causes more collision stuff move crazily which makes electrons hard to pass through. Thus, as T increases, R increases. R is also dependent on resistivity of material. Say like, copper is more conductive than nickle. 

V Python2: E Potential two straight vertical lines

from visual import *



k = 8.988e9


#left line positive charges
#Charge 1

c1 = sphere(pos=(-2,3,0), radius=0.2,color=color.blue)

q1= 1e-9   #charge of Charge 1

#charge2
c2 = sphere(pos=(-2,2,0), radius=0.2,color=color.blue)

q2= 1e-9   #charge of Charge 2
 #charge3
c3 = sphere(pos=(-2,1,0), radius=0.2,color=color.blue)

q3= 1e-9   #charge of Charge 3
#charge4
c4 = sphere(pos=(-2,0,0), radius=0.2,color=color.blue)

q4= 1e-9   #charge of Charge 4

#right line negative charges
#Charge 5

c5 = sphere(pos=(2,3,0), radius=0.2,color=color.red)

q5= -1e-9      #charge of charge 5

#charge6
c6 = sphere(pos=(2,2,0), radius=0.2,color=color.red)

q6= -1e-9   #charge of Charge 6
#charge7

c7 = sphere(pos=(2,1,0), radius=0.2,color=color.red)

q7= -1e-9   #charge of Charge 7
#charge8
c8= sphere(pos=(2,0,0), radius=0.2,color=color.red)

q8= -1e-9   #charge of Charge 8


#Observation Locations

loc1= sphere(pos=(0.5,2,0), radius = 0.1,color=color.green)

rc1= mag(loc1.pos-c1.pos)

rc2= mag(loc1.pos-c2.pos)
rc3= mag(loc1.pos-c3.pos)
rc4= mag(loc1.pos-c4.pos)
rc5= mag(loc1.pos-c5.pos)
rc6= mag(loc1.pos-c6.pos)
rc7= mag(loc1.pos-c7.pos)
rc8= mag(loc1.pos-c8.pos)



Vc1= k*q1/rc1


Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vc4=k*q4/rc4
Vc5=k*q5/rc5
Vc6=k*q6/rc6
Vc7=k*q7/rc7
Vc8=k*q8/rc8




Vnet= Vc1 + Vc2 + Vc3+Vc4+Vc5+Vc6+Vc7+Vc8

L1 = label(pos=loc1.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)



loc2= sphere(pos=(-0.5,1,0), radius = 0.1,color=color.green)

rc1= mag(loc1.pos-c1.pos)

rc2= mag(loc1.pos-c2.pos)
rc3= mag(loc1.pos-c3.pos)
rc4= mag(loc1.pos-c4.pos)
rc5= mag(loc1.pos-c5.pos)
rc6= mag(loc1.pos-c6.pos)
rc7= mag(loc1.pos-c7.pos)
rc8= mag(loc1.pos-c8.pos)



Vc1= k*q1/rc1


Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vc4=k*q4/rc4
Vc5=k*q5/rc5
Vc6=k*q6/rc6
Vc7=k*q7/rc7
Vc8=k*q8/rc8





Vnet= Vc1 + Vc2 + Vc3+Vc4+Vc5+Vc6+Vc7+Vc8

L2 = label(pos=loc2.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

V Python 1: E Potential surrounding like a circle centered with three point charges

from visual import *

k = 8.988e9

#Charge 1

c1 = sphere(pos=(0,0,0), radius=0.2,color=color.blue)

q1= -1e-9   #charge of Charge 1

#Charge 2

c2 = sphere(pos=(-0.5,-0.5,0), radius=0.2,color=color.red)

q2= 1e-9      #charge of charge 2

#charge3

c3 = sphere(pos=(0.5,-0.5,0), radius=0.2,color=color.yellow)

q3=1e-9      #charge of charge 3

#Observation Locations

#loc1

loc1= sphere(pos=(0,2,0), radius = 0.1,color=color.green)

rc1= mag(loc1.pos-c1.pos)

rc2= mag(loc1.pos-c2.pos)

rc3= mag(loc1.pos-c3.pos)

#rc4= mag(loc1.pos-c4.pos)

#rc5= mag(loc1.pos-c5.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L1 = label(pos=loc1.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

 #loc2

loc2= sphere(pos=(1.414,1.414,0), radius = 0.1,color=color.green)

rc1= mag(loc2.pos-c1.pos)

rc2= mag(loc2.pos-c2.pos)

rc3= mag(loc2.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L2 = label(pos=loc2.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

#loc 3

loc3= sphere(pos=(2,0,0), radius = 0.1,color=color.green)

rc1= mag(loc3.pos-c1.pos)

rc2= mag(loc3.pos-c2.pos)

rc3= mag(loc3.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L3 = label(pos=loc3.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

#loc4

loc4= sphere(pos=(1.414,-1.414,0), radius = 0.1,color=color.green)

rc1= mag(loc4.pos-c1.pos)

rc2= mag(loc4.pos-c2.pos)

rc3= mag(loc4.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L4 = label(pos=loc4.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

#loc5

loc5= sphere(pos=(0,-2,0), radius = 0.1,color=color.green)

rc1= mag(loc5.pos-c1.pos)

rc2= mag(loc5.pos-c2.pos)

rc3= mag(loc5.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L5 = label(pos=loc5.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

#loc6

loc6= sphere(pos=(-1.414,-1.414,0), radius = 0.1,color=color.green)

rc1= mag(loc6.pos-c1.pos)

rc2= mag(loc6.pos-c2.pos)

rc3= mag(loc6.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L6 = label(pos=loc6.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

#loc7

loc7= sphere(pos=(-2,0,0), radius = 0.1,color=color.green)

rc1= mag(loc7.pos-c1.pos)

rc2= mag(loc7.pos-c2.pos)

rc3= mag(loc7.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L7 = label(pos=loc7.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

#loc8

loc8= sphere(pos=(-1.414,1.414,0), radius = 0.1,color=color.green)

rc1= mag(loc8.pos-c1.pos)

rc2= mag(loc8.pos-c2.pos)

rc3= mag(loc8.pos-c3.pos)

Vc1= k*q1/rc1

Vc2=k*q2/rc2

Vc3=k*q3/rc3

Vnet= Vc1 + Vc2 + Vc3

L8 = label(pos=loc8.pos,xoffset=20,yoffset=20,text= "V=%1.2f" % Vnet)

lab11 March 31 Gauss'slaw

In this class, we explore the Gauss's Law. 

Question: What is the relationship between the net flux and the net charge enclosed by a two-dimensional “surface”?

Process: By using a java applet, we model the flux and the charge on computer and draw the diagram on whiteboard. By counting the flux, we find that total flux is proportional to the total charge. Then, we conclude that Electric Flux over a Closed Surface = Charge enclosed by the Surface divided by eo.

Conclusion:

Faraday cage

Each of the suspensions include a string passing over the top of the cylindrical screen with small rectangular chunks of aluminum foil, a few centimeters in size, attached to the ends of the string. Note that one end of each string is outside the cylinder and one end of the string is inside the cylinder. 

Question:In particular, when the cylinder is charged, which, if any, of the aluminum foils will move away from the cylinder, either inward or outward?

Process:Turn on the charge generator, the foil outside moves outward; The foil inside keeps still.

In this case, the black cylinder is an electrical conductor. We did the experiment to prove that "if there are no moving charges inside a conductor, the electric field in the conductor must be zero."

Excess charge

Question:If the conductor has excess charge and it can’t be inside the Gaussian surface according to Gauss’ law, then what’s the only place the charge can be?

Process:we drew the diagram to show where are the charges. 

Question: 

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 tree

2) Get out of the car and lie down flat on the ground.

3) Seek the highest point nearby and put up your umbrella

4) Seek the lowest point nearby preferably a ditch or ravine.

5) Stay in your car

Conclusion:A car is seemed as an electrical conductor. Staying inside is safer.

Professor Mason put light ball and CD inside a microwave. The light bulb lightens up and the CD is burned. We assumed the cd is an electrical conductor and has free charges on the surface. As it heat up, the electrons move rapidly and damaged the texture of the cd. 

Then, we use Gauss's Law and symmetrical charge distribution to solve electrical fields. 

Question:

compute the magnitude of the electric field at a distance r from the center of a uniformly charged sphere of radius R with a total charge of Q throughout its volume, where r < R 

Process:To use Gauss’ Law to find the electric field, you’ll need to know the charge enclosed inside a sphere/cynlinder of radius r/cross-section area radius in which charges are distributed uniformly through the larger sphere/cynlinder of radius R

We get q=(r^2/R^2)Q for cylindrical shell, and q=(r^3/R^3)Q for sphere.  

We substitute the equation E=q/e, and get the charge of outside cylinder shown in the whiteboard. 

Here is the analogy problem of the Gauss's law problem. We use the formula to calculate the gravitational field , and get g=9.8m/s^2