Capacitance

Capacitance

In this activity, we created capacitors using two sheets of aluminum foil, separation distance (provided by sheets of paper). We carefully cut two square pieces of aluminum foil and measured the area to be 0.0316 m^2 for each. We then measured the thickness of a single page. We did this by measuring the thickness of 280 pages (making sure not to include the cover sheets and dividing the total number of pages by 2 since a single sheet has two "pages" front and back) and dividing by amount of sheets. We calculated that a single sheet measured 6.357E-2 m.

We connected a multimeter to the two sheets of foil via alligator clips. The positive end on one of the aluminum sheets and the negative end on the other aluminum sheet. We separated the two sheets of foil by a single sheet of paper and measured the capacitance with the multimeter. We did this again for 2, 10, and 15 sheets of paper and recorded the capacitance. Then we folded the aluminum sheets such that they had half of their original surface area (0.0158 m^2) and measured the capacitance again for 1, 2, 10, and 15 sheets of paper separation distance, respectively.

 When collecting data for capacitance, we pressed down on the pages so that there would be the least separation distance possible. That is, that the only separation distance is the thickness of the paper sheets, not of air or deformation in the pages which will create a higher separation distance and thus data with greater inaccuracy.

Here is the data we collected from our trials. We took this data and used excel to plot a Capacitance vs. Separation Distance graph as shown below. The blue curve corresponds to the data taken from the original foil surface area and the orange curve corresponds to half of the original surface area.

 The following observations were found to be true from the data we collected.

Resistance in Circuits

Resistance in Parallel Circuits

We were given three 150 Ω resistors and wired them in parallel. Theoretically, the total resistance should be given by the formula in blue above. The theoretical resistance of the three 150 Ω resistors in parallel is 50 Ω. We took a multimeter and measure the resistance to be 49.4 Ω which is within 1% error.

Here we analyzed a circuit where some resistors are in parallel and some are in series. We simplified the circuit in steps by combining resistors. We created a symbolic equation of the total resistance of the circuit as shown in black on the bottom right of the picture above.

Using our new skills, we created a symbolic equation for a new circuit (above). We were given the resistance of each resistor. We found that the theoretical value for the total resistance in the circuit is 52.2 Ω.

We now took the resistors and wired them up according the schematic given in the previous picture. We measured the value to be 53.6 (although it fluctuated). We subtracted the internal resistance of the multimeter, 1.4 Ω, and found that the experimental value for total resistance was the same as the calculated value.

We found that resistors add directly when they are wired in series and add in inverse when wired in parallel. We also saw that it was easier to break up a circuit into simpler circuits when trying to obtain the total resistance for the circuit.

Testing the Loop using Kirchoff's Rule

Here we applied Kirchoff's Law to find the current at different points in the circuit, across the resistors. We ended up with three equations and three unknowns for the currents. We used a matrix to solve for the individual currents (in mA). The individual values for the currents as labeled are i_1 = 1.137 mA, i_2 = 0.999 mA, i_3 = 0.138 mA.

We then set up the circuit on a breadboard as shown above. We used a potentiometer as resistor #2 and adjusted it until the resistance was 2.15 kΩ (The potentiometer was very sensitive and it was very difficult to turn it to a value of exactly 2.00 kΩ).

Next, we measured the resistance across resisors R_1, R_2, R_3, the potential differences, and currents i_1, i_2, and i_3. The data is shown in the table below.

As we can see, the % discrepancy was incredibly large (130% for the third resistor!). There is a huge source of error in the potentiometer. Therefore, we ran the experiment again, except that this time we swapped the potentiometer with a resistor that had a measured resistance of 2.13 kΩ (shown in the picture below).

We took our new measurements as shown in the table below.

We can see that our new values were much more accurate than the previous ones. The largest sources of error were in the resistors and in the battery. The battery provided 1.45 V instead of 1.50 V, and the resistors did not all match the theoretical resistances that we had used to calculate our theoretical currents.

Electric Potential

Electric Potential

In this lab, we took conductive paper that had a line and dot painted on it with metallic paint. A voltage supplier created electric potential difference between the line and dot via alligator clips. This voltage was measured as 15.02 V (please disregard the reading on multimeter as the picture was from a previous, and failed, attempt at the experiment).

We then measured two points on the higher and lower voltage, respectively. When we measured two points on the higher voltage, the reading came out to be -0.62 V. When we measured two points on the lower voltage, we got a reading of 1.37.

Starting from the point on the right, we measured the potential difference at 1 cm intervals going towards the line. We used excel to record the data and to create a  Potential vs Position graph as shown below.

Immersion Heater

Immersion Heater

In this activity we took a 3.41 W immersion heater and submerged it in a given amount of water for 10 minutes. We were to calculate how much the change in temperature of the water would be after 10 minutes had elapsed. While we performed the calculations, the water was heated and the temperature change with respect to time was tracked using LoggerPro software. At the end we compared our experimental results with the results of the LoggerPro data. The change in temperature was indeed within the uncertainty range that we had calculated.

Ohm's Law, Relating Electric Potential, Current, and Resistance

Measuring Electric Current

Here we connected a battery to power a light bulb via alligator clips. We see that all the components are functional and the bulb lights up.

We connected an ammeter to a position before and after the bulb as shown in the diagram below to measure the electric current at those positions. We found that the current was 110 +/- 5 mA for both positions. This means that the current remains constant.

Measuring Electric Potential, Current and Resistance

(Picture of the resistor used)

Our next set up was a bit different and consisted of a voltage supplier and a resistor shown in the picture below. An ammeter was again used to measure the current and a voltmeter was used to measure electric potential.

We plotted a graph showing the current vs voltage for the data that we obtained from our resistor and the data that our neighboring group obtained from their resistor. The two lines were plotted and fitted with linear trendlines which suggests that there is a linear relationship between current (I) and voltage (V), (they are proportional). We can say that I=kV, where k is a constant. We learned that solving for k (k=I/V), we get the resistance (R). So now we can say that I=RV. From our graph, the slopes of the trendlines are actually the resistance of the resistors. The two resistors were different and had a different resistance. This is shown by the slopes of the two lines, they are both different.

Other Variables

In this part of the experiment, we measured the resistance of 7 different wires with unique characteristics. They varied in material, length, and diameter. Plotting a graph of resistance vs length we get the curve shown above. The majority of the points follow a positive linear slope, however two of the points make the trendline inaccurate for this graph. We found the following relationships to hold true following our analysis.

Area (A)  is inversely proportional to the resistance (R), the wire length (L) is directly proportional to the resistance, and the resistivity of the material (rho) is directly proportional to the resistance. The data point that does not follow the trendline in the resistance vs length graph has to do with it having a larger cross sectional area (larger diameter).

Electric Field Lines and Flux

Flux as a Function of Surface Angle

In this activity we find the flux of a surface as a function of its surface angle.

In the picture above we have a bed of nails, which represents an electric field and a metal square which we use to represent a surface. The number of nails (electric field lines) that would go through the surface is varied by adjusting the angle of the angle the surface makes from perpendicular to the electric field. We used a measured height to calculate the angle rather than using a protractor directly and using a calculated column on LoggerPro software to automatically calculate the corresponding angles with respect to the measured heights (h). The formula we used to calculate the angle in degrees (Ref Degrees) was (arcsin(h/hypotenuse))*(360/2pi). The hypotenuse was measured as 6.75 cm. The program automatically calculated for radians therefore we multiplied by the formula to get it in degrees. To get the angle perpendicular to the surface, we needed to adjust the degrees where the flux is negative. This was done using the following for when flux is negative: 90+(90-Ref Degrees). This adjustment help form the graph. We then went ahead and created another calculated column for this angle in radians. The formula was just degrees*(2pi/360). The number of field lines (flux) was plotted with their corresponding angle to observe the relationship between the two.

After plotting the points we added a trendline to the graph. The one that fit it most properly was a sine function (though in this case it has a horizontal shift factor that makes it a cosine). We can see from the graph that when the angle is 0, the flux is also at a maximum (49). When the angle is zero, the surface is perpendicular to the electric field and therefore the flux is at maximum, but when the angle is pi/2 then the flux is zero because the surface is parallel to the electric field. This same behavior is consistent with the cosine function: cos(0)=1, cos(pi/2)=0. The electric field multiplied by the cosine of the angle between it and the surface gives us the flux. Moreover, the cosine also defines a dot product between E vector and dA vector.

Electric Flux Activity

Throughout the simulations, we made some observations about flux as seen on these whiteboard pictures.

Electric Fields

ActivPhysics: Electric Field, Point Charge

To observe and gain a better understanding of electric fields, we used ActivPhysics to perform a few simulations involving point charges and electric fields. This board shows the answers to the six questions that were asked in the simulation. I will go over these questions one by one.

1)  In this part we observed the nature of electric charges and the electric fields that they cause. In this case, we are using two positive point charges.

The first observation to note is that the electric field points away from the charges. This means that two positive charges repel each other.

Next we observed that the closer the point charges were to each other, the magnitude of the electric field increased.

2) In this part, we set the first charge (q1) to +10.0E-8, the second charge (q2) to 4.0E-8, and the seperation distance was 100 cm. We calculated the electric field caused by q1 as shown on the whiteboard picture. The electric field was 900 N/C.

3)  In this simulation, we examined how an electric field changes as a point charge changes.

We saw that with a positive charge, the field lines point outward while with a negative charge they point inward. The lines seemed to start and end in the same place relative to the different charges we observed. However, we did notice that the larger the charge, the larger the line density of the field.

4,5,6)  In this simulation we observed a uniform field with two plates. We saw that the electric field was spaced uniformly and that the magnitude was the same between the planes and it was independent of the position. 

The Electric Field from an Extruded Charge Distribution

In this problem, there is a rod with uniform charge throughout. The charges of the rod are divided into ten 1.00 cm "point charge" segments. We are to calculate the electric field from the rod to two points in space: P which is located 5.00 cm to the left and on axis to the rod, and P' which is located 5.00 cm perpendicular the the center of the rod. We used the formula E=Q/(r^2) to find the electric field of each rod segment. We were given that the entire rod has a charge of 5.00E-8 C. The charge is divided into ten 1.00 cm pieces therefore we used a charge of 5.00E-9 C to calculate the electric field of each segment. We used excel as shown below to calculate the charge on each segment. The initial distance (r) was 0.055 m (point P to the center of the segment) and the distance increased by 0.01 m. We calculated the charges with excel for all of the segments and added them together to get the electric field of the rod on point P, which came out to be 5.96E4 N/C.

Electric Field from a Uniformly Charged Rod

Here we calculated the electric field on the point P' from the same rod as the previous part. In this part, the direct distance from the point to the center of a segment piece (r) is needed to calculate the electric field on the point. We know that the vertical distance from P' to the rod is 0.05 m. We then complete the triangle using the variable x as the horizontal distance from the point on the rod just mentioned to "r". Solving for the distance we get that r=/sqrt(0.05^2+x^2). Now we use our equation for electric field to get E=(2k(C/10)

Electric Field vs. Distance

Electric Field Hockey

In this game, we place charges on the screen to move a charged puck to a goal.