Day 15 - Capacitors

Our group performed an experiment in which we were to verify the relationship between distance and capacitance. We used an old physics textbook with varying degrees of separation based on the number of pages, as well as two sheets of aluminum. We used distances of 1, 10 and 20 pages of separation within the textbook. We obtained our capacitance values and graphed them, and found their to be an inversely proportional relationship between the distance and capacitance. As the distance increased, the capacitance decreased significantly. Our kappa value that was based on the first trial was .367, slightly off by a factor of 10 from the theoretical value of 3.5 for paper.











Later for the class experiment we were presented with two capacitors in which to measure the capacitance of. We found them to have 1.088 and 1.165 micro farads, respectively. We then proceeded to set them up in series and in parallel in order to measure the resulting capacitance values. By doing this we were able to verify the theoretical knowledge that we have been taught in class; that when capacitors are in parallel their capacitance values are simply added, and when they are in series their equivalent capacitance value is found in the same way that equivalent resistance of a parallel resistor is found, with the equation shown below.

Day 14 - Circuits in Series and Resistors

Lab Manual:

-Voltage measurements

Based on our measured values there was no voltage loss at all, since the initial and final values were the same. The current is also constant in the circuit.

-Parallel circuits

Based on what we measured both voltages remained constant in the parallel series resistors, and the initial and final currents are the same.

-Measuring resistors

We learned how to decode resistors by knowing which color bands stood for what value, and we could see if their color coding reflected their actual theoretical value by measuring them.


Equivalent resistance challenge problem:

Our group was given the series of resistors above and was asked to find the equivalent resistance. I was able to calculate it by doing each section individually and then breaking it down into several steps. We found the resistance to be 52.17 Ohm's. We then set up a real life version of this circuit, shown below:

The actual measured resistance of this system was 51.9 Ohm's which is very accurate considering we were using cheap 1/10th of a cent resistors.

Application of the loop rule several times

We were presented with the complex circuit above and found 3 equations in order to solve for the three I values. Eventually we were able to substitute the equations back into eachother to find alll the values, as shown below. Our values were actually incorrect because an arithmetic error was made at some point, since a calculator gave us values of 1.14, 0.99, and .14 amps, respectively.

In order to verify what we just got on the calculator, we set up a breadboard pictured below and measured the values which were 1.15, 1.005, and 0.148, respectively. In contrast these are very close to what we calculated theoretically.

Day 13 - Difference in Potential & Continuous Charge Distribution

Experimental lab demonstrating electric Potential

For this lab our power supply is rated at 15 volts, and the potential difference between the two pins was 15.54 Volts that we measured. As long as we know the charge, the change in voltage will always be known. If two points on the field were randomly chosen, the potential would be zero. Therefore we know that no work is required.

This data set shows the change in voltage based on the position, and the ratio for both of them when divided by one or the other. Each new position was in an increment of 1 centimeter. We wanted to know the work in order to move the given charge from certain distances:

A. 0 to 3

B. 4 to 6

C. 5 to 2

Resulting values -

Since we know that W=Q*delta(V), it is easy to calculate since we know the change in voltage going from each point based on the data table above.

THe potential versus position graph above shows that there is a directly proportional relationship between the two. This is similar to a concept learned in the previous 4A class.

Day 12 - electric potential

We were required to make the light bulbs as dim as possible with the given equipment. Ours was the dimmest out of the entire class such that it could almost not be seen. I was not sure if this set up would work initially, but jeff had the same idea as I did and we went through with it. It should be noted that Bianca did not come up with this idea.







Change in temperature by electric heating element.

For this experiment we used a coil heater with a length of 42 centimeters. We attached it to a battery of 4.5 volts and let it run for 10 minutes. We needed to find the change in temperature based on the length of the heater, as well as the time and voltage, which is calculated below:

We found the resistance of the heater by multiplying the power by the length, and then dividing by the area. We then used the resistance to find the current based on Ohm's law. The relative uncertainty was found by seeing the deviation in the resulting power by plugging in two different values. The power was easy to solve because we knew both the voltage and current. 

We then proceeded to perform the same experiment with twice the voltage, and the graph is shown below. The red line is 4.5 V and the blue line is 9V. For the 4.5V test the temperature changed from 23 to 25 celsius. For the 9V test it changed from 24.6 to 32.1, which was much more significant. The rate at which the heat increased was more than double, even though the value for the voltage was only doubled.

It can be proven theoretically why the temperature increased nearly exponentially This is because:

delta T = Pt/mc = (V^2*t)/(RMC)

We substituted the definition of power being equal to V^2 divided by R, and then when you plug in a value of V that is double than a previous initial value, it is increased more than two-fold because its value is squared after it has been doubled in respect to the temperature change equation.

Day 11 - Ohm's Law

Practical application of Ohm's law with lightbulb:

We needed to make a complete system where a light bulb would light up, with only one wire and a battery. We were able to accomplish this and in order to make it brighter we simply doubled the power supply.

For this next experiment we chose a resistor and connected it to an ammeter and power supply. We knew the voltage because we were able to control it from the supply. We continually increased the voltage for each test and the actual measured values for the voltage and current for our group and our neighboring group are given below.

The other group appeared to have a smaller resistor, so it would make sense that their current values would be higher than ours for similar voltage. It should be noted to focus on the measured values because their measured values were not the same as ours for each trial.

The graph above shows that there is a linear and directly proportional relationship between the voltage and current. It was also noted that the slope of the lines are actually the values of the resistors, which makes perfect sense. Our group's resistance was 28.3 while the other group's was 17.7. This makes sense now why they would have higher current values compared to us. 

Copper vs. Nickel-Silver

For the next series of tests we were given a series of coils that differed in diameter and length for the wires that were wrapped around them. Only one was copper while the rest were some combination of nickel and silver. We measured the resistance of each coil and the data is shown below. We found that the meter had an error of 1.9 ohm's because we corrected it by finding the actual resistance of the power meter itself.

It is evident that the resistance increases along with the length of the coil.

The graph shows a directly proportional relationship between the length and resistance. The effect that Area has on resistance is actually inverse, since we tested two coils with the same length but different diameters, and the resistance decreased with a larger diameter.

Day 10- Gauss' Law

Section 11.8 ActivPhysics - Gauss' Law

Question 1: Electric Field Outside a Charged Sphere
A solid sphere of 10-mm radius has +10 x 10^-10 C of electric charge distributed uniformly throughout the sphere. Use Gauss's law to determine the electric field caused by this charge at a distance of 15 mm from the center of the sphere. The answer is shown in the meter. If you have difficulties, the Advisor will help you in this first application of Gauss' law.

Question 2: Electric Field Outside a Charged Spherical Shell:
A 10-mm radius spherical shell, like a basketball, has +10 x 10^-10 C of electric charge distributed uniformly on its surface. Use Gauss' law to determine the electric field caused by this charge at a distance of 15 mm from the center of the sphere. After your prediction, you can click the "Shell" Object type and set R = 15 mm and compare the answer with your prediction. Consult the Advisor if you have difficulties.

Question 3: Electric field inside the charged spherical shell
Leave the Object type on "Shell." Use Gauss' law to predict the value of the electric field inside the spherical shell. If you want, you can calculate the magnitude of the electric field at R = 5 mm from the center of the 10-mm radius spherical shell. After your prediction, move the R slider to 5 mm to check your prediction. If you have difficulty understanding the result, consult the Advisor.

Question 4: Electric Field Inside a Charged Solid Sphere
Predict the value of the electric field 5.0 mm from the center of a 10-mm radius solid sphere. The sphere is uniformly charged with +10 x 10^-10 C of electric charge. (a) First, determine the electric charge inside a 5.0 mm radius Gaussian surface. (b) Then, use Gauss' law to determine the electric field 5.0 mm from the center of the charged sphere. When finished, you can check your result with the simulation. Consult the Advisor if you have difficulties.

Calculations done in class for #1-4

Question 5: Electric Field Inside a Charged Solid Sphere
Show that the electric field INSIDE the solid uniformly charged sphere varies as:

E = [k Q/(R_{charged sphere})²](R/R_{charged sphere})

where R_{charged sphere} is the radius of the charged sphere (10 mm in the simulation), Q is the total charge on the sphere (adjustable with the Q slider), and R is the distance from the center of the sphere at the position where the electric field is being calculated.

#5 Calculation:

E=(kQ/R^2)* (r/R)

Q= λ*V

E=(k λV)/r^2

 λ=Q/(Vr^2)

E=((K(Q/(V_r))*V)/r^2

v/V_r=r^3/R^3

     E = (KQV)/(V_r*r^2)

E= (KQ/r^3)(r^3/R^3)

   = (KQ/R^2) * (r/R)