Exercise 1 electric fields in this exercise, you will use a digital multimeter to collect voltage data to graph electric fields. procedure place the sheet of graph paper on a table and center the clear tray over the grid. attach the end of each jumper cable to a metal nut by clamping the free alligator clip onto it as shown in figure 4. the figure is a photo of an alligator clip attached to a metal nut. it is attached so that one jaw of the clip is inside the nut, against the threads, and the other jaw is on the outside. figure 4. attaching the alligator clip to the metal nut. place the two metal nut conductors in opposite ends of the clear tray. they should be approximately centered and about 2.5 cm away from the ends of the tray. position the battery holder with a 1.5v battery outside of and slightly away from the tray so it cannot get wet. attach the jumper cables from the two conductors to the battery holder, one to the positive terminal and the other to the negative terminal. fill the tray with sufficient water to just barely cover the conductors. set your dmm to measure voltage by moving the dial to dcv, and its range to a voltage equal to or higher than that of the 1.5v battery. attach the negative black lead from the dmm to the negative terminal of the battery holder. attach a jumper cable to the positive red lead that comes from the dmm. to the other end of the jumper cable attach the washer. see figure 5. the figure is a photo of the experiment setup. the two nuts are positioned in the clear tray, each connected to opposite terminals of the battery holder. one lead from the digital multimeter is connected to the metal washer, and the other lead is connected to the negative terminal of the battery holder. figure 5. experimental setup. with the dmm's positive red lead, touch each of the conductors in the tray and record your findings. touching the negative conductor in the tray should result in a zero volt reading, touching the positive conductor should result in a reading that is the same as the battery output, and touching a distance halfway between the conductors should record a voltage equal to approximately one-half the voltage of the battery. if it does not, stir a few grains of salt into the water in the tray. using the second sheet of graph paper, draw the conductors' locations and label them with the voltage readings of your voltmeter. place the positive red lead of the dmm in the water again and note the voltage reading. move the lead around in such a way that the voltage reading is kept at the same value. how far does this path go? sketch this pattern on your graph paper and label the line with the voltage you chose. move the positive lead along additional voltage value paths and similarly sketch their patterns on the graph paper until you have well mapped out the area between and around the conductors. with a color pen or pencil draw a point any place on your map to represent a moveable positive charge. predict the path it would take by drawing a line with your colored pen or pencil. see figure 6. the figure is a sample graph of experimental results. it shows a 4.5 v trace encircling the +5v electrode. the 4.0 v, 3.5 v, and 3.0 v traces form arcs which curve around the +5v electrode. the 2.5 v, 2.0 v, and 1.5 v traces form arcs which curve around the 0v electrode. the 1.0 v trace encircles the 0v electrode. figure 6. sample graph of experimental results. note: your results will look different from the sample since you will use a rectangular dish, place your conductors in different positions, and use a 1.5v battery, but it still illustrates the concept of electric field mapping. take a photo of the graph. upload the image into graph 1. optional: if time permits, repeat this experiment with differently shaped conductors and compare their electric field maps.
1. what generalizations can you make from this exploration?
2. where would a positive test charge have the least potential energy?
3. how much energy must you add to the system to move one electron 1 m in a direction along one of the equal potential lines?
4. if lightning strikes a tree 20 m away, would it be better to stand facing the tree, your back to the tree, or your side to the tree? assume your feet are a comfortable shoulder width apart. explain your answer.
Solved
Show answers
More tips
- F Food and Cooking How to salt lard?...
- W Work and Career Can Skill Alone Make You a Professional?...
- C Computers and Internet How to Top Up Your Skype Account Without Losing Money?...
- P Philosophy Unidentified Flying Object - What is the Nature of this Phenomenon?...
- F Family and Home Protect Your Home or Apartment from Pesky Ants...
- O Other What is a Disk Emulsifier and How Does it Work?...
- F Family and Home What does a newborn need?...
- F Family and Home Choosing the Right Car Seat for Your Child: Tips and Recommendations...
- F Food and Cooking How to Get Reconfirmation of Registration?...
Answers on questions: Physics
- H Health Which of the following is true of research on insight? a) researchers have found that only human beings are capable of insight learning. b) researchers have found support for...
- M Mathematics A rectangular prism is cut along a diagonal on each face to create two triangular prisms. The distance between A and B is 5 inches. A rectangular prism with the horizontal...
- M Mathematics Seth purchases a video game for $62, what was the markup rate if the store purchased the game for $40?...
Ответ:
The answers are in the explanation section below
Explanation:
1) The generalization that can be made from the exploration is that as we move away from the positive electrode, the potential energy gets lower. If we move away from the negative electrode, then the potential energy becomes higher.
2) The positive test charge will have the least potential energy when it gets to the negative electrode point.
3) To move one electron 1m in a direction along one of the equal potential lines, there is no energy needed. Zero work will be required for a charge to move on the equipotential line.
4) If lightning strikes a tree 20m away, it would be better to face the tree or have our back facing the tree. This is because the equipotential line will be present at the point where our body stands, this will protect from electric shock.
The pattern to be sketched is attached.
Ответ:
x = 2.19 mm
I = 3.03E-9
Explanation:
Sin θ = λ/a = 0.600µm / 380µm = 0.00158
θ = sin¯¹ (0.00158)
θ = 0.090°
x = Rtanθ = 1.4 tan(0.090)
x = 2.19 mm
Approximating small amgle will be fine in this situation. By using small angle approximation, θ here will be 0.045°
Therefore sinθ = 0.785×10¯³
Using the formula
β = 2πa sin θ/λ
β = 2×3.142×380µm×0.785×10¯³/0.600µm
β = 3.124
So,
β/2 = 1.562
Utilizing above value in the formula
I = Io (sin(β/2)/(β/2))²
I = (10×10¯6) (0.0272/1.562)²
I = 3.03E-9