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Unformatted text preview: Scott Hughes 3 February 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 2: Electric fields & flux; Gausss law 2.1 The electric field Last time, we learned that Coulombs law plus the principle of superposition allowed us to write down fairly simple formulas for the force that a swarm of N charges exerts on a test charge Q : ~ F = N X i =1 Qq i r i r 2 i . We found another simple formula for the force that a blob of charged material with charge density exerts on a test charge: ~ F = Z QdV r i r 2 i . And, of course, similar formulas exist if the charge is smeared out over a surface with surface density , or over a line with line density . In all of these cases, the force ends up proportional to the test charge Q . We might as well factor it out. Doing so, we define the electric field : ~ E ~ F Q = N X i =1 q i r i r 2 i ( N Point charges) = Z dV r i r 2 i (Charge continuum) . Given an electric field, we can figure out how much force will be exerted on some point charge q in the obvious way: ~ F = q ~ E . In cgs units, the electric field is measured in units of dynes/esu, or esu/cm 2 , or stat volts/cm all of these choices are equivalent. (Well learn about statvolts in about a week.) In SI units, first of all, we must multiply all of the above formulae by 1 / 4 . The units of electric field are then given in Newtons/Coulomb, or Coulomb/meter 2 , or Volts/meter. Lets look at a couple of examples. 2.1.1 Example: Electric field at center of charged ring Take a ring of radius R with uniform charge per unit length = Q/ 2 R . What is the electric field right at its center? Well, if we dont think about it too carefully, we want to do something like this: ~ E = Z r r 2 ds, 11 where the integral will be take around the ring; ds will be a path length element on the ring, r is the radius of the ring, and r is a unit vector pointing from each length element on the ring to the rings center ... But wait a minute! You should quickly notice that for each contribution due to a little length element on the ring, there is an equal contribution from the rings opposite side pointing in the opposite direction . Each contribution to the ~ E field at the center of the ring is precisely canceled by its opposite contribution: So the field at the center of the ring is zero. 2.1.2 Example: Electric field on rings axis Suppose we move along the rings axis to some coordinate z above the disks center. The field does not cancel in this case, at least not entirely: Considering the contribution of pieces of the ring, we see that there is a component in the radial direction, and a component along z . The radial component cancels by the same symmetry argument as we used above. The z component, however, definitely does not cancel: Radius R Height above ring z cancels in radial direction, does not cancel in z direction....
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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.
 Spring '10
 ScottHughes

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