Open the electric force simulation.
Electric field and electric force:
Consider charge Q1 = 10.0 x 10-8 C in the simulation as the source of an electric field, just as the earth's mass is the source of a gravitational field. This electric field exerts an electric force on any charge, such as Q2, that is placed in the field (just as the earth's gravitational field exerts a force on the mass of your body when placed in that field).
Consider charge Q1 = 10.0 x 10-8 C in the simulation as the source of an electric field, just as the earth's mass is the source of a gravitational field. This electric field exerts an electric force on any charge, such as Q2, that is placed in the field (just as the earth's gravitational field exerts a force on the mass of your body when placed in that field).
We can probe the field produced by source charge Q1 by using a positive test charge Q2. The field that the source charge Q1 produces at some position points in the direction of the force
that Q1 exerts on Q2, divided by Q2:
that Q1 exerts on Q2, divided by Q2:
.Suppose that Ball 1 at the bottom of the simulation screen is the source of the electric field. Increase the charge of Q2 to +4.0 x 10-8 C. Use it to probe the electric field produced by Q1. To see this field, grab Q2 and move it around. Notice that if Q2 gets closer to the location of Q1, the magnitude of the electric force exerted on Q2 increases--the electric field is greater. If Q2 is moved farther away from Q1, the magnitude of the electric force exerted on Q2decreases--the electric field is less. After systemmatically moving Q2 around the region surrounding Q1, summarize in words your observations about the direction of the electric field produced by Q1 at different points in space and how the field's magnitude varies. When finished, compare your observations to those of the Advisor.
Answer: As seen from the above picture the electric field of the two charges point away from each other. This makes sense because the two positive charges are repelling each other. As they were moved closer together the electric field increased in strength and decreased when spread further apart.
Question 2: Field due to single positive charge
Set Q1 = +10.0 x 10-8 C. Use Q2 = +4.0 x 10-8 C as the test charge to measure the electric field due to source charge Q1. Move Q2 to a point 1.0 m to the right side of the source charge Q1 (when r12 = 100 cm--2.5 divisions on the screen). Observe the direction of the electric force exerted by Q1 on Q2. Use the force shown in the simulation
to calculate the electric field caused by Q1 at that point. Then, compare your field calculation with that of the Advisor.
Set Q1 = +10.0 x 10-8 C. Use Q2 = +4.0 x 10-8 C as the test charge to measure the electric field due to source charge Q1. Move Q2 to a point 1.0 m to the right side of the source charge Q1 (when r12 = 100 cm--2.5 divisions on the screen). Observe the direction of the electric force exerted by Q1 on Q2. Use the force shown in the simulation
to calculate the electric field caused by Q1 at that point. Then, compare your field calculation with that of the Advisor.
Answer: Using the given force(F12) and charge(q2) we were able to determine the magnitude of the electric field to be 900 N/C.
Open the electric field representation simulation.
Question 3: Representing an electric field
The traditional method for representing an electric field is shown for a single point charge. You can press the pointer down on the screen to see the magnitude of the electric field
at a point. Vary the value and the sign of the charge that is the source of the field (use only one point charge for this activity) and develop some rules for the way that the electric field is represented. Consider in particular:
The traditional method for representing an electric field is shown for a single point charge. You can press the pointer down on the screen to see the magnitude of the electric field
at a point. Vary the value and the sign of the charge that is the source of the field (use only one point charge for this activity) and develop some rules for the way that the electric field is represented. Consider in particular:
• the direction of the lines;
• where the lines start or end relative to the sign of the charge causing the field;
• the separation of the lines in a particular region relative to the magnitude of the field in that region;
• the magnitude of the field on a line and in the dark region next to the line, and
• the number of field lines that eminate from or terminate on a point charge relative to the magnitude and sign of that charge.
When finished, compare your observations to those of the Advisor.
(Negative Charge)
Answer: Based on the two above pictures we can see that when the charge is positive the electric field points away from it and when the charge is negative the electric field points towards it. Another conclusion we can see from these pictures is that as the electric charge increases (positive or negative) the density of the electric field also increases.
Open the electric field representation simulation.
Force on a charge in the electric field
The force 
that an electric field 
exerts on a charge q in that field is:
that an electric field exerts on a charge q in that field is:
.
In this simulation, the source charges that produce the field are positive charges distributed uniformly on one metal plate and an equal number of negative charges on the other plate. The electric field produced by these charges is represented by the green lines between the plates. (Be sure that the "Equipotential lines" button is off.) The pointer can grab the charge q and move it to different places between the plates. Before moving the charge, answer these questions.
Question 4: Uniform Field
Explain why the word "uniform" was used in describing the electric field in the middle region between the plates. When finished, compare your thinking to that of the Advisor.
Explain why the word "uniform" was used in describing the electric field in the middle region between the plates. When finished, compare your thinking to that of the Advisor.
Question 5: Force on a charge in a uniform field
Adjust the plate charge per unit area to 0.6 x 10-8 C/m2 and the value of the charge q between the plates to +0.5 x 10-8 C. Suppose you place +q in the middle near the top plate . Note the arrow showing the force on q and the magnitude of that force. Is the force greater in magnitude, the same, or less if qis moved directly below its present position so that it is near the bottom plate? Justify your choice. After your prediction, move q down and compare the force on it when in this new position. When finished, compare your thinking to that of the Advisor.
Adjust the plate charge per unit area to 0.6 x 10-8 C/m2 and the value of the charge q between the plates to +0.5 x 10-8 C. Suppose you place +q in the middle near the top plate . Note the arrow showing the force on q and the magnitude of that force. Is the force greater in magnitude, the same, or less if qis moved directly below its present position so that it is near the bottom plate? Justify your choice. After your prediction, move q down and compare the force on it when in this new position. When finished, compare your thinking to that of the Advisor.
Answer: The force at the top of the plate and the bottom of the plate turned out to be the same. This fits with what we expected to happen because the electric field between the two plates is uniformly distributed and spaced. This means that the electric field does not depend on its location.
Question 6: Force on a negative charge
If you leave q in the middle half way between the plates and change the sign of q from +0.5 x 10-8 C to -0.5 x 10-8 C, what happens to the force exerted by the field on q. After your prediction, adjust the charge of q to check your thinking. Try different magnitudes of q and different signs. Move q around to different places between the plates and observe the direction and magnitude of the force. Then, move the charge to the sides of the plates where the electric field is not uniform. Are your observations consistent with the equation that relates the field and the force:
If you leave q in the middle half way between the plates and change the sign of q from +0.5 x 10-8 C to -0.5 x 10-8 C, what happens to the force exerted by the field on q. After your prediction, adjust the charge of q to check your thinking. Try different magnitudes of q and different signs. Move q around to different places between the plates and observe the direction and magnitude of the force. Then, move the charge to the sides of the plates where the electric field is not uniform. Are your observations consistent with the equation that relates the field and the force:
Answer: The force stays the same whether the charge is positive or negative.
Using excel we determined the Electric Field from Two Point Charges at varying distances. We used this table to help determine the electric field from the two point charges at the measured distances.
Using the distances from the table above you can easily calculate the electric field to each point.
E-Field
Vectors from a Uniformly Charged Rod
Given the above data we determined the electric field from each element to the indicated point. Our answers were calculated in excel and we found the following:
Based on these results we determined the total electric field for a parallel axis to be 5.965·10^5 N/C. The total electric field (Y component) for a perpendicular axis is 1.273·10^6 N/C.
Electric Field Hockey
The goal of the game was to successfully score the positive puck into the goal with the minimum amount of positive or negative charges. For level one I was able to score the goal with only 2 charges.
The second level proved much more difficult but after several tries I was able to score the puck with only two positive charges.
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