Download How To Find the Electric Field for a Continuous Distribution of Charges

How To Find the Electric Field for a Continuous
Distribution of Charges
For a continuous distribution of charge, it’s really the same thing as for
point charges, except that you treat the continuous distribution as if it is a
bunch of infinitesimally small point charges added together. So the total field
i
~ tot = Pni=1 k∆q
r̂,
at X, which is the sum of fields due to each piece ∆qi , E
ri2
R kdq
~ tot =
becomes an integral E
r̂ (note that it’s a vector integral). To
r2
actually calculate this integral, the trick is to convert dq into some quantity
you can integrate over.
1. First, think about the dimensionality and symmetry of the charge distribution. Is it a 1D (linear), 2D (surface) or 3D (volume) system?
Does it have a circular symmetry (in which case, polar coordinates
might be best) ? If 3D, does it have spherical or cylindrical symmetry?
The answers to these questions will help you in step 8. Make a diagram and define appropriate coordinates, if they are not given in the
problem.
2. Identify the spot X where you want to find the field.
3. Mentally divide the continuous charge distribution up into a bunch of
tiny pieces which are equivalent to point charges. Exactly how to do
this depends on the shape of the object. For example, if you have a
line charge, cut it up into little pieces dq along the line.
~ i at X which is the contribution
4. For each piece dq, draw the vector dE
~ is away
to the electric field due to dq, remembering that direction of E
from positive charges and towards negative ones. When considering
the field due to dq, ignore the presence of any other charges.
5. Find the magnitude of this contribution to the field at X: dE = kdq/r2 ,
where r is the distance from the source charge dq to the spot X you’re
at.
~ into components in your chosen coordinate
6. Decompose the fields dE
system.
~ i vectorially, i.e. add up the î, ĵ and k̂ compo7. Now add the fields dE
nents. Note that you can often use symmetry to simplify your answer.
The sum in a particular direction will
be an integral which is the sum
R kdq
of contributions of the type Etot = r2 .
8. To evaluate this integral, you need to convert dq into some space coordinates you can integrate over, using what you know about the charge
density distribution. Here is where the answers to the questions in step
1 will help you.
a. 1D case: dq = λ(x)dx. λ is the linear charge density, which is
charge per unit length.
b. 2D case: dq = σ(x, y)dA. σ is area charge density, which is
charge per unit area.
c. 3D case: dq = ρ(x, y, z)dV . ρ is volume charge density, which
is charge per unit volume.
A careful choice of coordinates based on the geometry of the charge distribution will make all the difference here. Polar coordinates (r, θ) may
be convenient for circular geometries; cylindrical coordinates (for which
dV = rdrdzdφ) or spherical coordinates (for which dV = r2 sin θdrdθdφ)
are usually best for systems with cylindrical or spherical symmetry.1
Note that for cases when charge density is uniform (i.e. λ, σ or ρ are
constants), you may be able to evaluate it from total charge Q, and
total length, area or volume.
9. Substitute the relation between charge and space coordinate from step
8 into the integral(s) from step 7.
10. Write the distance r between dq and X in the integral(s) in terms of
your space coordinates.
11. Crank through the integral(s) to get the answer.
12. Check your dimensions! Electric field has dimensions of force per
charge; from F = ma you can remember that force has dimensions
of mass × length/time2 .
1
Note r here is a space coordinate, not the distance between dq and X. Take care to
make a diagram and label each relevant quantity carefully.