For a charged ring when we find its electric field ox the (x,y,z) components, $E_x$ and $E_y$ vanish and remains the $E_z$ component.
a) Please explain in your own words why the $E_x$ and $E_y$ components are equal to zero and if you can predict this without using mathematics.
b) Can you tell the why an electron moves if released far away from the ring on the z axis?
c) What if the electron is released a bit off z-axis, on the x direction?
d).For a spherically symmetric uniform charged shell, what are the locations where the $E$ field is zero if we consider the $r$, $theta$ and $phi$ components?
See the figure. Because of the symmetry of the arrangement (ring) with respect to z axis, the other components of the electric field E along axes x, and y cancels each other.. Only the component of the field along z will remain. This is predictable without math, just by drawing a picture of the experiment setup.
If the ring is positively charged (as in the above figure), the electron will be attracted towards the center of the ring along the z axis. Once in the center of the ring it will continue its trajectory until the electric force will break its acquired speed. Then it will start to oscillate back and forth along the z axis.
If the electron is released a bit off the z axis, there will be a nonzero force action on it along the other axis (x for example), so that it will be more attracted toward a side of a ring and it will end up “sitting” on that side of the ring.
There will be only the component of the Filed along the coordinate for which the arrangement is symmetric. Since a sphere is symmetric over any of its diameters, there will be present (nonzero) just the r component of the electric field.The other ($theta$, and $phi$) components of the field will be zero.
(To be more clear the field will be directed exactly along the radius.)