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Ch 21: Electric Charge and Electric Field

Chapter 21, Problem 21

Point charges q1 = -4.5 nC and q2 = +4.5 nC are separated by 3.1 mm, forming an electric dipole. (b) The charges are in a uniform electric field whose direction makes an angle of 36.9° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.2 * 10^-9 N•m?

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Welcome back everybody. I'm going to go ahead and draw some electric field lines here. Let me make this just a little bit greater. And what we are going to do is we are going to set up a di pole moment through these electric field lines and we're told a couple of different things here. So let me just go ahead and are dai pole moment. We'll get rid of this last line here. Great. So we are told that we're going to have a charge on each end connected by some non conducting material. Right? And we are told that the charge on on each of these is going to be plus or minus 3.2 nano columns. We are also told that the angle at which the disciple makes with the Electric field is going to be 40°. We are told that the distance between the two charges is 4. millimeters. We are told that the torque that is being felt by the dipole moment in the electric field is 6.3 times 10 to the negative ninth Newton meters. And we are tasked with finding what is the magnitude of the electric field of our di pole moment. Now we need to find e and we have a couple other terms. I'm gonna start out with a formula that relates some terms that we have. So we know that for the torque on the die pole, this is just going to be equal to the dipole moment, times the magnitude of the electric field, times the sine of theta. We know what he is and we know what data is but we don't know what our di pole moment is. Well, there's another formula for that are dipole moment is simply just the magnitude of one of the charges. The so essentially the absolute value of one of the charges times the distance between them. So I'm gonna go ahead and sub this in right here and this gives us that our torque is equal to Q. D. E. Sign of data. Now we need to isolate the magnitude of the electric field by itself. So I'm gonna divide both sides by Q. D. Signed data. You d sign data as all of these terms cancel out. And we get that the magnitude of our electric field is equal to the torque on the die pole divided by the magnitude of the charges, times the distance between them, times the sine of the angle data. Now that we have our formula, let's go ahead and plug in our values. We have that E. Is equal to 6.3 times 10 to the negative ninth. Newton meters divided by our charge, which we will want in normal. Cool. Um So I'm gonna multiply this by 10 to the negative nine to convert two columns, extend my fractional sign there. Um Then we are going to multiply by the distance between them which we need in meters. So I'm gonna multiply by 10 to the negative third to get meters and then times the sine of our angle. You're going to see that the meters cancel out, leaving us with our desired units of newtons per Coolum and specifically the magnitude of the electric field. When you plug this into your calculator is 7 29 newtons per giving us a final answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.