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Sketch the electric field between the two conducting plates shown in Figure 18.49, given the top plate is positive and an equal amount of negative charge is on the bottom plate. Be certain to indicate the distribution of charge on the plates.
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<b>Figure 18.49</b>
Figure 18.49
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OpenStax College Physics Solution, Chapter 18, Problem 39 (Problems & Exercises) (4:51)

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This is College Physics Answers with Shaun Dychko. This top conductor has some positive charge put on it and an equal amount of negative charge is put on this bottom conductor and you can see that the charges are not evenly distributed; the end of this top conductor that is closer to the bottom conductor has a greater density of positive charge here and likewise, there's a greater density of negative charge here on the bottom conductor where it's closer to the top conductor. Now this is because these charges are attracting each other and whereas over here where the conductors are far apart, the charges are spaced apart a long ways because all the excess charge has been squished over here in this end where the conductors are close together. Okay! Now the only sort of non-obvious part above the electric field lines is this curve that they have here. Now remember that to figure out what direction the electric field should go, you have to imagine a positive test charge somewhere and ask yourself what direction would the force be on that positive test charge? Well a positive test charge here would be attracted to the negative charges on the bottom conductor so there's gonna be a sort of downward component to the electric field. They will also be repelled by the charges in the top conductor but the charges on the left side of this test charge have a greater density—there's more positive charges here—and so there's gonna be a greater force directed to the right of repulsion from these charges compared with this low density distribution of charges over here which will be repelling this test charge a little bit to the left and so this difference in the charge on the left side and on the right side of this test charge accounts for this sort of... horizontal net component to the force that this test charge would experience and therefore the electric field is gonna have that same direction somewhat to the right. So if we could say that this is the repulsion from the charges that are on the left side and this is the repulsion from the charges on the right side— this one's shorter— and there's gonna be a net sort of down like this due to the repulsion so that's a bit to the right from vertical and then there's gonna be attraction downwards as well but the repulsion from these positive charges is going to be greater than the attraction to the negative charges over here and so the electric field will be pointing somewhat to the right and down. Now consider what happens when the test charge gets here; now we have a greater attraction to the leftmost negative charges here and a smaller attraction to the right charges over here and this is gonna have a net sort of direction to the left of vertical and that's what we are seeing here in this field line curvature is it's a bit to the left here. Okay! So the same analysis applies all the way along here; this one I drew straight down here but I suppose I should even add another field line bulging a bit to the left here like this because if you put a test charge here, it will experience repulsion from these positive charges but there's no charge to the left of it that would exert a force with some component to the right like there is for these ones and so this line is gonna bulge out like this. And likewise if this charge was down here, it's going to experience an attraction to these negative charges here but there's nothing over here to attract it that way and so it's gonna have its component really towards the right here trying to get towards these negative charges in the bottom conductor.