lec03 - Scott Hughes 8 February 2005 Massachusetts...

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Unformatted text preview: Scott Hughes 8 February 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 3: Electric field energy. Potential; Gradient. Gauss’s law revisited; divergence. 3.1 Energy in the field Suppose we take our spherical shell of radius r , with charge per unit area σ = q/ 4 πr 2 , and squeeze it. How much work does it take to do this squeezing? To answer this, we need to know how much pressure is being exerted by the shell’s electric field on itself. Pressure is force/area; σ is charge/area; force is charge times electric field. Hence, our intuitive guess is that the pressure should be P guess = (electric field)(charge/area) = 4 πσ 2 . (We have used the fact that the electric field just at the surface of the shell is 4 πσ .) This is almost right, but contains a serious flaw. To understand the flaw, note that the electric field we used here, E = 4 πσ , includes contributions from all of the charge on the sphere — including the charge that is being acted upon by the field. We must therefore be overcounting in some sense — a chunk of charge on the sphere cannot act upon itself. It’s not too hard to fix this up. Imagine pulling a little circular disk out of the sphere: The correct formula for the pressure will be given by P = (electric field in hole)(charge/area on disk) = (electric field in hole) σ . We just need to figure out the electric field in the hole. Fortunately, there’s a clever argument that spares us the pain of a hideous integral: we know that the field of the hole plus the field of the disk gives us the field of the sphere: E hole + E disk = E sphere = 4 πσ Just outside the sphere = Just inside the sphere. 23 What’s E disk ? Fortunately, we only need to know this very close to the disk itself: from the calculations we did in Lecture 2, we have E disk = +2 πσ Outer side of disk =- 2 πσ Inner side of disk. We’re now done: E hole = 2 πσ (Pointing radially outward) which means that the pressure is P = 2 πσ 2 . (Purcell motivates the factor of 1 / 2 difference by pointing out that one gets the pressure by averaging the field at r over its discontinuity — the field is 0 at radius r- ² , it is 4 πσ at r + ² . This argument is totally equivalent, and is in fact based upon Purcell problem 1.29.) Now that we know the pressure it is easy to compute the work it takes to squeeze: dW = F dr = P 4 πr 2 dr = 2 πσ 2 dV . On the last line, we’ve used the fact that when we compress the shell by dr , we have squeezed out a volume dV = 4 πr 2 dr . Now, using E = 4 πσ , we can write this formula in terms of the electric field: dW = E 2 8 π dV . The quantity E 2 / 8 π is the energy density of the electric field itself. (In SI units, we would have found ² E 2 / 2.) It shows up in this work calculation because when we squeeze the shell we are creating new electric field: the shell from r- dr to r suddenly contains field, whereas it did not before. We had to put dW = (energy density)...
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This note was uploaded on 07/19/2010 for the course 8 8.022 taught by Professor Scotthughes during the Spring '10 term at MIT.

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lec03 - Scott Hughes 8 February 2005 Massachusetts...

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