Required math: calculus, algebra
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 2.41, 2.42, 2.43, 2.44.
Here are a few more examples of the calculation of electric field and potential.
Example 1. Given a square sheet of charge with side length and surface charge density , find the electric field at height above the centre of the sheet.
From Example 3 in this post, the field at distance above the centre of a square loop of charge of side length and linear charge density is
If we think of this square loop as an element of the square sheet, where the thickness of the loop is (so that the overall square has side length ) then and the field of the sheet is
The mathematical software (Maple, in this case) needs a bit of help with this integral, so we can use the substitution ; to get
The field points vertically upwards, by symmetry.
Example 2. If the electric field is given by
( and are constants), find the charge density.
In this case, we can use the differential form of Gauss’s law: . In spherical coordinates, the divergence for a vector field with its component equal to zero is
Example 3. In a uniformly charged solid sphere, find the force of repulsion between two hemispheres.
To make the problem definite, we’ll consider the ‘north’ and ‘south’ hemispheres, so the problem becomes one of finding the vertical force between these two hemispheres.
From Example 2 in a previous post, the electric field inside a uniformly charged sphere with density is
The vertical force on a volume element is therefore
So the total repulsive force on a hemisphere is
Note that we can’t just find the repulsive force between two hemispheres of a spherical shell and then integrate over shells of sizes from zero up to the radius of the sphere because this doesn’t take into account the force between hemispheres of different sizes.
Problem 4. A northern hemispherical shell of radius has a uniform surface charge density of . Find the potential difference between the north pole and the centre of the base of the hemisphere.
For an inverted hemispherical bowl, we can use the method of Example 5 in this previous post, except the limits on the integral are now 0 to . We therefore get
At the north pole, so
At the centre, so we need to take a limit. We can rewrite the formula as
The lowest order term in the expansion of the square root is
so the limit as is
and the potential difference is