Spherical Coordinates 2

Question

Fundementals of applied electromagnetics i am wondering where does that formula come from? (that i signed in pic.)

Answer 1

These are well-known properties of vectors, so if you are having trouble recognizing them, I really suggest a review. But here are the essentials:

At any point on a surface where the surface is smooth, there will be a plane tangent to the surface, and a line that normal to that plane (and therefore, to the surface).

So if you are given a vector $\mathbf E$ at that point, you can project $\mathbf E$ on to the normal line, to get the "normal component" $\mathbf E_n$, and you can also project it onto the tangent plane to get the "tangent component" $\mathbf E_t$. And because the normal line and tangent plane are orthogonal to each other,

$$\mathbf E = \mathbf E_t + \mathbf E_n$$

Now, if the direction of the normal line is given by the unit vector $\mathbf{\hat n}$, then the signed length of the projection of $\mathbf E$ is given by the inner product $\bf E\cdot \hat n$. And since lines are one dimensional, every vector on the line is a multiple of $\mathbf{\hat n}$. So it must be that $$\mathbf E_n = (\mathbf E\cdot \mathbf{\hat n})\mathbf{\hat n}$$

And therefore, the tangential component is given by

$$\mathbf E_t = \mathbf E - \mathbf E_n = \mathbf E - (\mathbf E\cdot \mathbf{\hat n})\mathbf{\hat n}$$

For a cylinder, the unit normal vector to the surface is the radial vector $\mathbf{\hat r}$, because the tangent to a circle is perpendicular to the radius.

For a sphere, the unit normal vector to the surface is the radial vector $\hat {\boldsymbol\rho}$, for exactly the same reason.