Part 2 Goals: Calculate the electric field (vector!) that is...

Part 2 Goals: Calculate the electric field (vector!) that is produced by more than one point charge. Two equal but opposite point charges are located on the x-axis: +Q at x = -a and -Q at x = +a. The goal is to find the electric field created by these two charges at a point P on the y-axis at y = +a. Draw some axes and these charges in the space below. Mark the point P. 1. Draw the setup, and then draw two vectors representing the electric field at P created by these two charges. The arrows should be drawn with their tails at P. 2. How do the magnitudes of the two electric field vectors compare? Explain why your answer makes sense. 3. What will be the direction of the net electric field at P? Explain briefly why your answer makes sense. 4. Write down expressions for the x- and y-components of the electric field due to the negative charge. Check that the signs of these components make sense. 5. Determine the net electric field (and, that's a vector... ) at point P, in terms of Q, a, and relevant constants.

Transcript

0:00 Hello everyone.

00:01 In this problem, we're asked to find out a bunch of things about an electric field that is a combination of, or the result in the electric field of two point charges that are equal in magnitude, but opposite in sign.

00:17 So we're asked to draw a diagram where these are located such that the positive charge with charge plus q is located at x equals minus a, and the charge with negative charge minus a.

00:30 Is located at x equals plus a.

00:32 And we're interested in finding the electric field, the net electric field at point p, which is along the y -axis at point p, which is also a distance a away from the origin.

00:45 So this is the diagram here that you should have in one.

00:49 So that's what we're interested in, right? we drew the point p up here.

00:54 And now the negative, sorry, the positive chart is located along the negative, x -axis.

01:02 And so when we draw the electric field coming from this charge, we just connect the point where the positive charge is located to the point p up here.

01:12 And the electric field is going to point along this line.

01:16 And it's going to point away from the positive charge because remember that the electric field is defined such that it describes the direction in which a positive test charge would move.

01:27 So if you placed a positive charge at point p, you would want to move away from the positive charge located at minus x equals minus a.

01:35 Now the opposite happens when we're considering a charge minus q located at x equals plus a.

01:43 And that's because if i placed a positive charge at point p, it would be attracted towards this point charge minus q, right? because if it's a positive charge, then it's going to be attracted to the negative charge.

01:54 And so the electric field that is coming from the negative charge, charge to point p is going to be an attractive electric field that points towards minus q from the point of p.

02:10 So that's what we have there.

02:12 Now, the second question, our second part is asking, how do the magnitudes compare to each other? how does the magnitude of e plus and e minus compare to each other? while the magnitude of the electric field for a point charge is just q over 4 pi epsilon not times r squared.

02:26 So it's given by this formula right over here.

02:30 And so you can see that the distance between positive q and negative q to the point b is going to be the same.

02:39 Right.

02:40 So you know that this side or each of the sides of these right -handed triangles here.

02:46 Right.

02:46 So the y component and the x component of these right -handed triangles are both a.

02:51 So the distance is going to be actually when we work out later, it's going to be this r -minus, which is also equal to r -plus, by the way.

03:00 So this is equal to the square root of 2 times a.

03:03 So the distance is the same, and all other parameters are also the same...

Question

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