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.

Electrostatic Foces

Interactions of Scotch Tape Strips

In this activity we observe the electrostatic forces in charged pieces of scotch tape. 

In this part, we placed two pieces of tape, sticky side down, on the table and pulled them off. We then brought the non-sticky sides of tape together. The two pieces of tape repelled each other. The closer the tapes got to each other, the more the repulsion force was.

For this part, we placed two new pieces of tape on the table. Next we placed two other pieces of tape on top of these ones. We pulled the pairs of tape of the table and then pulled the tapes apart. When the two top strips were brought toward one another, they repelled. The same thing happened when the two bottom strips were brought together. However, when a top and bottom strip came together they had an attractive force.

From our observations, we can conclude that there are two different types of charges. Two objects with the same charge will have a repulsive force, just as witnessed when the two "top" or two "bottom" tapes were brought into close proximity to one another in our experiment. Also, we observed that two objects with opposite forces will attract each other, such as when a "top" tape came into proximity with a "bottom" tape.

Electric Force Law Video Analysis Activity

In this activity, we observe the repulsive electric force of two like-charged balls. One ball is suspended by a string while another is secured to a stick and brought into proximity of the hanging ball as seen in the video below.

We drew a free body diagram to show the forces acting on the ball in the video. There is a tension from the string (T), the force due to the mass of the ball and acceleration of gravity (mg), and an electrical force (F_e). We were able to separate the forces into their components and consequently isolate the electric force in terms of mass (M), earth's gravitational constant (g), separation distance (X), and length of the string (L). Since M, g, and L were given, our variable X was easily obtained using the video and LoggerPro software.

We used LoggerPro software to track the motion of both balls from the video. We also used the software to create a graph showing Force vs. Separation Distance. Separation distance (X) was calculated as the difference of the position of the two balls. We can see that the points on the graph fit closely to the power trendline. The equation of the curve was y = (2.085x10^-5)x^(-1.842).

Here we answered some questions in conclusion to our experiment. We were able to show that  the electric force was roughly inversely proportional to the square of the distance (x^2) between the charges. We saw this when we added a power trendline to our graph. From the trendline curve equation, we found the power to be -1.842 which is very close to the -2.000 result we would like to have.

The percent difference was obtained by using the equation shown in the picture (2a). We took the absolute value of the difference between the true value and experimental value. The whole quantity was divided by the true value and then multiplied by 100% to obtain a percentage. It is important to note that the picture contains mistakes. The 1.793 above was obtained from an earlier incorrect trendline. The trendline was corrected and the percent error was actually only 5%. 

From this experiment, it is not possible to tell what the sign of the charge on either ball is. We know that both repel each other which means that they have the same charge. The only way to discover the charge on the ball is to introduce another object with a known charge (for example, + charge). If the new objects would repel, then the ball is of the same charge (+) and if they attract then the ball is (-).

We noticed that there is some uncertainty in our experiment. The uncertainty comes from various sources. However, we believe that the main sources came from the tracking of the objects on LoggerPro. The initial scaling of the video ("x" inches on the screen = 1 m) also had some level of uncertainty. Our points did not all fit within the trendline but the percent error was still within a reasonable range therefore we feel comfortable with our results.

Heat Engines - Diesel Cycle, Brayton Cycle

Diesel Cycle

Brayton Cycle