Derivation of potential of two closed curves s and s'

In summary: Now, we can rewrite dx' dx + dy' dy + dz' dz as dr' ⋅ dr, where dr = (dx, dy, dz) and dr' = (dx', dy', dz'). This gives us:cos ε = (x' - x)dr' ⋅ dr / √[(x' - x)^2 + (y' - y)^2 + (z' - z)^2] ||ds' ⋅ ds||Finally, we can substitute this into our original formula:
  • #1
faheemahmed6000
18
0
I am reading Maxwell's "a treatise on electricity and magnetism" and I came across a formula of "potential of two closed curves s and s' "

##M= \iint\dfrac{cos\varepsilon}{r} dsds'##
where:

##M=## potential of two closed curves s and s'

##\varepsilon=## angle between elements ds and ds'

##r=## distance between elements ds and ds'

s and s' are two closed curves.

The book has a derivation of this formula in "article 423" but its extremely hard to understand because of lack of vector notation.

Can someone give me an easy to understand derivation of this formula using modern vector notation?
 
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  • #2


Hello there! As a fellow scientist, I would be happy to assist you with understanding this formula in modern vector notation. Before we begin, let's break down the components of the formula:

- M represents the potential of two closed curves s and s'. This can also be thought of as the potential energy between the two curves.
- The integral sign ∫ indicates that we are summing up the contributions of all the elements ds and ds' along the curves s and s'.
- The quantity cos ε represents the cosine of the angle between the elements ds and ds'.
- The term 1/r represents the inverse of the distance between the elements ds and ds'.

Now, let's start with the derivation. We will use vector notation to make things easier to understand.

First, let's define the position vectors of the elements ds and ds' as r and r', respectively. In vector notation, we can write this as:

r = (x,y,z) and r' = (x',y',z')

Next, let's define the vector between the two elements as r' - r. This gives us the distance between the elements as:

r' - r = (x' - x, y' - y, z' - z)

Now, we can use the dot product to find the angle between the two elements:

cos ε = (r' - r) ⋅ (ds' ⋅ ds) / ||r' - r|| ||ds' ⋅ ds||

where || || represents the magnitude of the vector.

Note that the vector ds' ⋅ ds represents the direction of the elements, and ||ds' ⋅ ds|| represents the length of the elements.

Using the definition of the dot product, we can rewrite this as:

cos ε = (x' - x)(dx' dx + dy' dy + dz' dz) / ||r' - r|| ||ds' ⋅ ds||

Now, let's consider the distance between the elements, which is represented by ||r' - r||. Using the Pythagorean theorem, we can write this as:

||r' - r|| = √[(x' - x)^2 + (y' - y)^2 + (z' - z)^2]

Substituting this into our equation for cos ε, we get:

cos ε = (x' - x)(dx' dx + dy' dy + dz
 

Related to Derivation of potential of two closed curves s and s'

1. What is the significance of deriving the potential of two closed curves s and s'?

The potential of two closed curves s and s' can be used to calculate the electric potential at any point in space due to the charged curves. This is an important tool in understanding and predicting the behavior of electric fields.

2. How is the potential of two closed curves s and s' calculated?

The potential of two closed curves s and s' is calculated by integrating the electric field over the area enclosed by the curves. This can be expressed mathematically as: V = ∫E · dA.

3. What is the relationship between the potential of two closed curves s and s'?

The potential of two closed curves s and s' is directly proportional to the charge enclosed by each curve. This means that as the charge of the curves increases, the potential also increases.

4. Can the potential of two closed curves s and s' be negative?

Yes, the potential of two closed curves s and s' can be negative. This occurs when the electric field is directed towards the interior of the curves, resulting in a negative potential value.

5. How is the potential of two closed curves s and s' affected by the distance between them?

The potential of two closed curves s and s' decreases as the distance between them increases. This is due to the inverse square law, where the electric field strength decreases with the square of the distance.

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