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Lec03 - Scott Hughes Massachusetts Institute of Technology...

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Scott Hughes 10 February 2004 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 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 to write down this equation. This answer turns out to be almost right — we just have to divide by two: P = 2 πσ 2 . Purcell gives a careful derivation of where this 1 / 2 comes from. Physically, it arises because we need to average the field at r over its discontinuity — the field is 0 at radius r - ² , it is 4 πσ at r + ² . 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 0 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) dV amount of work into the system to make that field.
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3.2 Electric potential difference I move a charge q around in an electric field ~ E . The work done by me as I move this charge from position ~a to ~ b is given by U ba = Z ~ b ~a ~ F me · d~s = - Z ~ b ~a ~ F q · d~s = - q Z ~ b ~a ~ E · d~s . ~ F me is the force that I exert. By Newton’s 3rd, this is equal and opposite to the force F q that the electric field exerts on the charge. This is why we flip the sign going from line 1 to line 2. Make sure you understand this logic — that sign flip is an important part of stuff we’re about to define! On the last line, we’ve expressed this in terms of the electric field. Recall that one of the things we learned about electric fields is that they generate conservative forces. This means that this integral must be independent of the path we take from ~a to ~ b . We’ll return to this point later. Since the work we do is proportional to the charge, we might as well divide it out. Doing so, we define the electric potential difference between ~a and ~ b : φ ba = - Z ~ b ~a ~ E · d~s .
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