Question

A disk of charge is an idealization of many physical devices: a capacitor plate, a small patch of any surface etc. In this problem you will find the electric field of such a (ideal) disk.

a) Consider first a ring of radius R with a total charge +q uniformly distributed on it. Calculate the electric field along the axis of the ring (i.e. along the line per- pendicular to the plane of the ring and through its center). Show that your solution makes sense for distances along the axis which are much larger than the radius R.
b) Consider a charge-Q constrained to slide along the axis of the ring. Show that the charge will execute a simple harmonic motion for small displacements per- pendicular to the plane of the ring..
c) Consider now a dise of radius Ro and a uniform surface charge density o. Use the result from part a) to calculate the electric field along the axis of the disc (i.e. along the line perpendicular to the plane of the disc and through its center). Consider all components of the electric field.
d) Check the units of your result for the electric field in part c) and show that the limits z ≪ Ro and z ≫ Ro make sense.

Answer 1

To find the electric field along the axis of a ring with uniformly distributed charge, we can calculate the electric field due to each infinitesimal charge element and sum them up. For distances along the axis much larger than the radius of the ring, the ring can be approximated as a disk, and the electric field will be the same as the electric field due to a uniformly charged disk. The electric field due to a uniformly charged disk along its axis can be calculated using the surface charge density and the distance from the center of the disk.

Step-by-step explanation:

To find the electric field along the axis of a ring of radius R with a total charge +q uniformly distributed on it, we can break the ring into infinitesimally small charge elements and calculate the electric field contribution of each element at the point along the axis. The electric field due to a point charge q at a distance r is given by E = k * q / r^2, where k is the electrostatic constant.

For distances along the axis that are much larger than the radius R, the small elements of the ring can be approximated as point charges, and the electric field at each point on the axis can be treated as the sum of the electric fields due to all the point charges. This means that the electric field along the axis of the ring will be independent of the radius R and will be the same as the electric field due to a uniformly charged disk.

To calculate the electric field due to a uniformly charged disk of radius R and surface charge density σ at a point along its axis, we can use the result from the previous part. The electric field at a point on the axis of the disk is given by E = σ / (2ε0) * (1 - z / √(z^2 + R^2)), where ε0 is the permittivity of free space and z is the distance of the point from the center of the disk along the axis.

When z ≪ R, the term z / √(z^2 + R^2) becomes negligible and the expression for the electric field simplifies to E ≈ σ / (2ε0). This makes sense because when the distance from the center is much smaller than the radius, the disk will appear as a flat surface and the electric field will be constant and independent of the distance.

When z ≫ R, the term 1 - z / √(z^2 + R^2) approaches 1 and the expression for the electric field simplifies to E ≈ σ / (2ε0) * (1 - 1) = 0. This makes sense because when the distance from the center is much larger than the radius, the disk will appear as a point charge and the electric field will decrease to zero at large distances.