**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

Using

we get

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

AkHello!!

your blog is very nice! It will be even more useful if you could attach a small photo describing the geometry of system you are working on, in example 4

gwrowePost authorJust picture a sphere like the Earth, then cut it in half at the equator. We’re finding the potential difference between the north pole and the centre of the Earth in the northern half of the sphere.

tygHello! I’m trying to work example 1. For lambda I’m getting an additional term: (sigma * da)/2 + ((da)^2) / 4a. If that’s clear. I basically found the area to be (a+da) * (a+ da) – a*a, multiplied by sigma, then divided all of that by 4a to get lambda.

Here, do you just say that da squared is close enough to zero that you don’t need the term?

gwrowePost authorRemember that is a

linearcharge density, that is, the charge per unitlengthof a wire. In this example, we’re building a wire by cutting a very thin strip of width from an area, where thesurfacecharge density of the area is , that is, has units of charge per unitarea. Thus is the amount of charge in a unit length of this strip. The area of a unit length of the strip is so the amount of charge in that unit length is . Another way of looking at it is that we’re looking for the amount of charge contained in an increment of area where is the area of a unit length of a strip with width , so that .TomI wonder, do you know a way to work out the integral without using mathematical software?

gwrowePost authorThe integral in equation 3 can be done by substitution. Let

Now let