# Component of electric field tangential to spherical shell is zero

For a spherical shell of radius R with a static uniform surface charge density, the electric field component of $\theta$ and $\phi$ is zero.

The reason supplied by my notes is this:

"Due to the uniform nature of the charge distribution on the surface of the sphere, that part of electric field tangential to the surface must be zero".

How can I 'see' that the $\phi$ and $\theta$ component is tangential to the surface of any given spherical shell?

It pretty much spells it out for you in the quote, you could think of it in the same way as gravity works on earth. If you stand up straight, you will not be pulled any way except down by gravity, you won't fall left or right, or backwards or forwards, if you are on level ground.

Look up central force on Wikipedia, with regard to spherical symmetrical potentials.

Just to finish off, in a Cartesian coordinate system, x, y and z are orthogonal, (at right angles to each other), in spherical coordinates, $r$, $\theta$ and $\phi$ are orthogonal.

This does only hold if the spherical coordinate system is centred at the centre of the spherical shell. If you take e.g. two spherical shells, this will not be true for both of them in the same coordinate system. For one spherical shell with the coordinate system centred at its centre, spherical coordinates give a natural parametrisation with $r=R_{shell}$ and $\vartheta, \varphi$ being variable. Thus it is clear that the vector components along the angles constitute tangential components to the shell.

If you wish to calculate to electrical force at sompe point $\vec{R}$ outside the sphere, you have to integrate over the sphere the field due to a surface element.

A surface Element is given by - $$dS=r^{2}\sin(\theta)d\theta d\phi$$

The element of charge on it is - $$dq=\sigma dS$$

So an element of the electric fiels is - $$d\vec{E}=k\frac{dq}{\vec{R}^2}\hat{R}$$

where $\hat{R}=\frac{\vec{R}}{|\vec{R}|}$

now we have to intgrate - $$\vec{E}=\intop_{S}d\vec{E}=k\intop_{S}\frac{\sigma}{|\vec{R}|^3}\vec{R}dS=k\intop_{0}^{\pi}d\theta \intop_{0}^{2\pi}d\phi \frac{\sigma}{|\vec{R}|^3}\vec{R}r^{2}\sin(\theta)d\theta d\phi$$

And you can see the due to the symetrical nature of the problem, the field you get does not depend on lies on the line that connects $\vec{R}$ and the center of the sphere. Now you may ask, what will happen if the center of the sphere is in an arbitrary loation in space? In the new system of coordinates the filed will not be radial to the center of the new coordiants, but this it not important as long a the sphere is the only source of electrical field. You can simply change coordinates and get to the above result. The physics of the problem does not depend on the location of the sphere, so the field is always radial to the sphere.