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

Heat Engines, Heat Capacity of Gas at Constant Pressure

Analyzing the Cycle

A graph of pressure (P) versus volume (V) was sketched for a heat engine. The heat engine has a gas that is heated. The gas expands and does work on the surroundings. Then it is then cooled at constant volume, using a stopper so the piston cannot move, reducing the pressure. The stopper is then removed and the volume decreases to its original position at constant pressure. The gas is heated at constant volume and the pressure rises to the original point and the process is repeated. The first three points were given and we found the rest of the graph information according to the other information given.

From point 1 to 2 the gas expands and volume increases, doing positive work (W+) on the surroundings. The pressure remains constant in this step. There is no heat in this process.

From 2 to 3 the volume is maintained constant and the gas is cooled. Since the volume remains constant, there is no work done. however there is heat loss to the surroundings in this process (Q-).

From 3 to 4 the volume decreases while the pressure remains constant. There is work done on the system (W-) by the surroundings but there is no heat entering or leaving the system.

From 4 to 1 the volume is kept constant while the pressure increases. There is no work done but positive heat (Q+) enters the system.

The internal energy was calculated at each of the four points using the formula E_int = (3/2)PV. We included these energies (in Joules) in our sketch next to each point in red.

The Carnot Engine Cycle

Here we see a repeating process that includes isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic expansion.

We calculate the internal energy at each point by using the equation E_int = (3/2)PV. We are to calculate the heat and work for each of the parts of the carnot cycle (pending).

ActivPhysics Simulations

In this first simulation (problem 6) we are calculating the heat capacity of a gas at constant pressure. We assume 1 mole of gas is in the cylinder.

 The temperature in this simulation starts at 200K with a constant pressure of 166 kPa and volume of 10 dm^3.

We let the simulation run until the temperature reaches 401K. At this point we can see the changes in volume, temperature, heat, work, and internal energy. We divide the change in heat (Q_f - Q_i) by the change in temperature (T_f - T_i) and by the 1 mole (n) of gas in the cylinder: (Q_f - Q_i)/((T_f - T_i)(n)). This gives us a 20.9 J/(mol K).

We ran the simulation from the previous point and stopped it immediately. This gave us a heat change of 354 J and a temperature change of 18 K. Using the same formula in the previous calculation, we calculated the specific heat of the gas at constant pressure and found it to be 19.7 J/(mol K).

Again we ran the simulation and stopped it. We calculated for heat capacity again and got 20.9 J/(mol K).

We repeated the same steps one last time and got the heat capacity of the gas at constant pressure to be 20.8 J/(mol K).

Averaging the values for the four trials gives us 20.6 J/(mol K). This comes very close to the actual value 20.4 J/(mol K).

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In this simulation (problem 7), we started the simulation at an initial temperature of 200K and stopped it at 797K.

We did the same as the in the previous calculations for heat capacity of a gas at constant temperature. This time we got 20.8 J/(mol K). This matches the 20.8 J/(mol K) we were supposed to get.

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On this whiteboard, we derive the specific heat of an ideal gas at constant pressure (problem 8). We see that we get the same 20.8 J/(mol K) that we obtained from the simulations in the previous problem.

The Fire Syringe, State Variables and Ideal Gas Law

The Fire Syringe

In this experiment, we have an air-tight cylinder with a piston that allows for a variation in volume. We want to know the temperature of the gas (air) inside the cylinder after it is compressed quickly. We derive an equation for the final temperature as shown in the picture below.

We made all of the necessary measurements and calculations to find the initial temperature volume. We used a thermometer inside the cylinder to measure the temperature of the air. Next we used a caliper to measure the diameter (from which we derived the radius) and the length of the chamber before compression. The volume was calculated using V=(pi)(L)(R)^2. We placed a small piece of cotton at the bottom of the chamber and quickly decreased the volume by pushing on the chamber. The video below shows this action at in slow motion.

The cotton inside ignited instantly and left a small black ring around the inside surface of the cylinder. This marked the final length (L) needed in calculating the final volume. We used the same caliper to measure this length and calculated our final volume. With this information, we were then able to calculate the final temperature of the gas inside the cylinder at the final volume.

Our calculated final temperature came out to be 792.5 K which is 967.1 *F. Compared to Ray Bradbury's novel Fahrenheit 451 (where the title refers to the temperature that paper will ignite), we are not surprised that the cotton mass caught fire. 

Since the volumes were calculated values, I needed to propagate for the uncertainties of initial and final volumes, respectively. I found the initial volume to have an uncertainty of 182 mm^3 and the final volume to have an uncertainty of 42.1 mm^3. These values were used in the next step, when propagating for the uncertainty in temperature.

Using the uncertainty for initial temperature to be half a degree Celsius or half a Kelvin, and using the uncertainties for volume from the previous step, I was able to propagate for uncertainty in the final temperature. The uncertainty cam out to be 113 K. So our final experimental temperature is 793 K +/- 113K.

ActivPhysics: State Variables and Ideal Gas Law

The picture above shows the answers and solutions for the ActivPhysics assignment we conducted in class. The assignment consisted of interactive simulations on the ActivPhysics website. We learned about the relationships between the state variables in the ideal gas law. The problems consisted of situations where certain variables were kept constant and we needed to solve for a pressure of volume.

Volume vs. Temperature for a Gas (Charles' Law I)

 Volume vs. Temperature Relationship

In this lab, we find the relationship between volume and temperature. We use a syringe connected air-tight to the opening of a flask to measure changes in volume. Submerging the flask into a large beaker of water with different temperatures will cause different volume changes in the volume of the flask-syringe system.

The first thing we needed to find was the volume of the flask. To do this, we measured the mass of the empty flask (with stopper and tubing). Next we filled the flask and tubing with water and measured the mass again. The difference between these masses gives us the mass of the water. Now, since we know the mass of the water and the density of water, we can rewrite the density formula and solve for volume. This gives tells us that the volume inside the flask and tubing is 139.2 cubic centimeters (cc).

In this lab, we find the relationship between volume and temperature. We use a syringe connected air-tight to the opening of a flask to measure changes in volume. Submerging the flask into a large beaker of water with different temperatures will cause different volume changes in the volume of the flask-syringe system.

Now we finish our set-up by connecting the syringe to the flask. We fill a beaker with water and connect the thermometer for data collection. We also obtain ice and hot water so that we can vary the temperatures of the water in the beaker.

The flask is submerged at a cold temperature and the syringe plunger begins to move down. The gas inside the system begins to cool and since pressure is constant, the volume goes down.

Three measurements were taken for temperature (T in *C) and for volume of the syringe (V in cc). All of the other columns in this chart were calculated. The temperature Kelvin was taken by adding 273.15 to the temperature Celsius measure. The volume of the flask was measured as explained earlier. The total volume was calculated by adding the volume of the flask plus the volume of the syringe once it had stopped moving (once the volume was stabilized). 

Using the table created, we constructed a graph that models Volume vs. Temperature. Excel was used to plot points, add a trendline, and find the equation of the line. From this graph, it is evident that the relationship between volume and temperature is linear. In other words, volume is proportional to temperature.
However, volume does not equal temperature. The temperature is multiplied by a coefficient which can be found by the slope of the line. The coefficient is 0.6494 cm^3/K, or in SI units, 6.494 X 10^-7 m^3/K.

Thermal Expansion, Latent Heats, and Relationships Between Pressure, Volume, and Temperature

Thermal Expansion

The top diagram shows that when a round metal with a circle cut out is heated, the metal expands and the hole gets bigger.

The bottom diagram shows a bimetallic strip. When heated, one metal expands moroe than the other metals and causes the metal to curve. This is because different metals have different expansion rates. This concept was confirmed in a lab experiment conducted by Prof. Mason.

Thermal Expansion of a Metal Rod

A metal rod has one end fixed in position while the other is free to move with expansion. A pulley supports the free end of the rod. As the rod expands, the pulley rotates and the change in length can be found by measuring the angle that the pulley moves through. With the help of the diameter, circumference, and angle of motion of the pulley, we can then use the original length of the rod to find the coefficient of thermal expansion of the metal (alpha). We found this coefficient to be 2.93 X 10^-5.

However, within every measured value there is a certain degree of uncertainty. To account for this, we propagated for the uncertainty of alpha considering the uncertainties in the measurements of temperature, diameter, and angle movement (theta). We propagated for the uncertainty in alpha and found that uncertainty was +/- 4.26 X 10^-6
This now means that alpha is 2.93 X 10^-5  +/-  4.26 X 10^-6.
We checked a thermal expansion table of metals and compared it with our results. We found several metals that were within the value and uncertainty range we came up with. However, we can make an educated guess and be fairly confident that the rod material was aluminum since it is cheap and it does not look like lead or glass, etc.

Latent Heat of Vaporization

In this experiment, we set up logger pro to track a temperature change in water as heat is added to it by a 308 W immersion heater. The initial mass of the water and cup was 150 g. We started the data collection and we can see the temperature rising in the graph.

At this point, the water is boiling and we can see from the graph below that the temperature keeps constant at 100*C.

We allowed the water to boil long enough for sufficient mass to be lost by steam. The final mass measured was 133.0 g, which means there was 17.7 g mass loss due to steam. However, the amount actually lost is greater because steam mass was being lost between the end of the data collection to the time when we measured the mass.
We measured the change in time when water was boiling by selecting the region where the line of the graph was horizontal at 100*C. This came out to be 116.5 s.

Now that we have the change in time, power, and mass we can solve for the latent heat of vaporization by doing the following.
First we took that the amount of energy is proportional to the power and time (W=Pt). This gave us that 35882 J of energy were used in the 116.5 s.
Next we took the amount of energy used and divided it by the amount of mass lost due to steam. This gave us the latent heat of vaporization for water to be 2027 J/g or 2.027 x 10^6 J/kg.

We took the results of experimental latent heat of vaporization for water and put them on a spreadsheet. With this we were able to calculate the standard deviation for the experiments. The standard deviation was 8.58x10^5 J/kg. We can take this to be the uncertainty in the experiment. This means that our value for latent heat of vaporization was 2.027 x 10^6 J/kg +/- 8.58x10^5 J/kg. This range of uncertainty covers the true value therefore we conclude that the our experimental value is accurate.

Latent Heats

In this problem, we will determine the final temperature of a 790 g block of ice that has 215 kJ of heat added to it. The total heat is equal to the heat required to take the temperature up from -5 *C to 0 *C, then the latent heat of fusion for water, and then the heat required to take the temperature of liquid water up to its final temperature.
We calculated that it takes 8255.5 J to rise the temperature 5 C* to 0 *C. Then we calculated that the latent heat of fusion was 1777500 J. This means that to change the temperature of solid water from -5 *C to liquid water at 0 *C, it takes 1785.76 kJ (82.555 kJ + 1777.5 kJ) of energy. This means 215 kJ is not enough to melt the block ice, therefore, the final temperature is 0 *C.
We also summed the energies and came up with a negative result. The negative implies that there needs to be more energy applied to the ice.

Relationships Between Pressure, Volume, and Temperature

This shows the relationship between pressure (P) and volume (V). We can see that the pressure and volume are inversely proportional. It is somewhat intuitive since if volume gets closer and closer to zero, the pressure must be approaching infinity. So P=A/V where A is a constant.

A demonstration with hot and cold water was conducted with a pressure sensor. The graph below shows the relationships between temperature and pressure.

This shows the relationship between pressure (P) and temperature (T). The pressure is proportional to the temperature. The graph of P vs. T is linear as seen from the demonstration.