22.8 Electric field of a continuous charge distribution by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
vCalc Companion Formulas | |
vCalc Formulary | 22.8 Electric field of a continuous charge distribution |
`dq=lambdadz` | `dq` |
`lambda=Q/L` | Charge per unit length |
`r=d-z` | `r` |
`E_z=(kQ)/L(1/(d-L/2)-1/(d+L/2))` | Total field along axis |
ab / Example 16.
Charge really comes in discrete chunks, but often it is mathematically convenient to treat a set of charges as if they were like a continuous fluid spread throughout a region of space. For example, a charged metal ball will have charge spread nearly uniformly all over its surface, and in for most purposes it will make sense to ignore the fact that this uniformity is broken at the atomic level. The electric field made by such a continuous charge distribution is the sum of the fields created by every part of it. If we let the “parts” become infinitesimally small, we have a sum of an infinite number of infinitesimal numbers, which is an integral. If it was a discrete sum, we would have a total electric field in the `x` direction that was the sum of all the `x` components of the individual fields, and similarly we'd have sums for the `y` and `z` components. In the continuous case, we have three integrals.
? A rod of length `L` has charge `Q` spread uniformly along it. Find the electric field at a point a distance `d` from the center of the rod, along the rod's axis. (This is different from examples 4 on p. 627 and 14 on p. 639, both because the point is on the axis of the rod and because the rod is of finite length.)
? This is a one-dimensional situation, so we really only need to do a single integral representing the total field along the axis. We imagine breaking the rod down into short pieces of length `dz`, each with charge `dq`. Since charge is uniformly spread along the rod, we have `dq=lambdadz`, where `lambda=Q/L` (Greek lambda) is the charge per unit length, in units of coulombs per meter. Since the pieces are infinitesimally short, we can treat them as point charges and use the expression `kdq"/"r^2` for their contributions to the field, where `r=d-z` is the distance from the charge at `z` to the point in which we are interested.
`E_z=?(kdq)/r^2`
`=?_(?L"/"2)^(+L"/"2)(k\lambdadz)/r^2`
`=k??_(?L"/"2)^(+L"/"2)dz/(d?z)^2`
The integral can be looked up in a table, or reduced to an elementary form by substituting a new variable for `d?z`. The result is
`E_z=k?(1/(d?z))_(?L"/"2)^(+L"/"2)`
`=(kQ)/L(1/(d-L/2)-1/(d+L/2))`.
For large values of `d`, this expression gets smaller for two reasons: (1) the denominators of the fractions become large, and (2) the two fractions become nearly the same, and tend to cancel out. This makes sense, since the field should get weaker as we get farther away from the charge. In fact, the field at large distances must approach `kQ"/"d^2`, since from a great distance, the rod looks like a point.
It's also interesting to note that the field becomes infinite at the ends of the rod, but is not infinite on the interior of the rod. Can you explain physically why this happens?
22.8 Electric field of a continuous charge distribution by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.