# Lec02 - Scott Hughes Massachusetts Institute of Technology...

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Scott Hughes 5 February 2004 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Lecture 2: Electric fields & flux; Gauss’s law 2.1 The electric field Last time, we learned that Coulomb’s 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 Q q i ˆ r i r 2 i . We found another simple formula for the force that a blob of charged material with charge density ρ exerts on a test charge: ~ F = Z Q ρ dV ˆ r i r 2 i . And, of course, similar formulas exist if the charge is smeared out over a surface with surface density σ , or over a line with line density λ . 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 ˆ r i r 2 i ( N Point charges) = Z ρ dV ˆ 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. (We’ll learn about statvolts in about a week.) In SI units, first of all, we must multiply all of the above formulae by 1 / 4 π² 0 . The units of electric field are then given in Newtons/Coulomb, or Coulomb/meter 2 , or Volts/meter. Let’s 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 λ = Q/ 2 πR . What is the electric field right at its center? Well, if we don’t think about it too carefully, we want to do something like this: ~ E = Z λ ˆ r r 2 ds ,

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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 ˆ r is a unit vector pointing from each length element on the ring to the ring’s 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 ring’s 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 ring’s axis Suppose we move off along the ring’s axis to some coordinate z above the disk’s 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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