Week 2: Continuous Charge and Gauss’s Law
- Continuous Charge
Charge distributions can often be continuous. We therefore define the followingcharge densi-
ties:
ρ =
dq
dV
σ =
dq
dA
λ =
dq
dL
for the charge per unit volume, per unit area, and per unit length respectively.
- Superposition Principle
To find the electrostatic field produced by a continuous charge density distribution, we use the
superposition principle inintegralform:
E~(~r) =k
∫ ρ(~r
0 )·(~r−~r 0 )d^3 r 0
|~r−~r 0 |^3
wheredV 0 =d^3 r 0 is the “volume element” – the volume of an infinitesimal chunk of the charge
in the charge distribution located at~r 0.
Because one has to integrate over the differentialvectors, this integral is remarkably difficult
to perform. We’ll revisit it in a much simpler form when we get to electrostaticpotential, a
scalar quantity that one can usually integrate more easily without this complication.
There are two more ways of writing this for the other two kinds of charge distribution:
E~(~r) =k
∫ σ(~r
0 )·(~r−~r 0 )d^2 r 0
|~r−~r 0 |^3
E~(~r) =k
∫
λ(~r 0 )·(~r−~r 0 )dr 0
|~r−~r 0 |^3
where in all cases the integral is over the entire charge distributionin question. Note that
dA 0 =d^2 r 0 anddL 0 =dr 0 are the “area element” and “length element” one uses in an
infinitesimal chunk of the distribution in the last two expressions.
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