Calculate the electric field above a ring of charge of radius `R` and charge density `lambda` at a point `P` centered on the axis of the ring.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

In general, the contribution to the total electric field `E` from an infinitesimal piece of charge `dE` is


The total field is then calculated by integrating over all the charge:


Where `k=1/(4pi epsilon_0)` and `r` is the distance from `dq` to the point `P`

For this problem, if the point `P` is a distance `z` above the ring then we can set up a triangle to get a relationship between `z` and the distance `r`.


The vector `dE` pointing at point `P` from a chunk of charge, can be broken up into two components, one in the radial direction and one in the vertical direction.


When considering the total electric field contribution from the ring, the radial contribution of every piece of charge will sum to zero and only the `z` contribution will be left due to the symmetry of the ring.


`E_z=int(k*dq)/r^2 *(z/r)`

`E_z=int (k*z*dq)/r^3`

`lambda` is the piece of charge divided by a piece of length along the ring.


From the fact that `r*theta` is equal to the arc length around a circle we can say that



`E_z=int (k*z*lambda dl)/r^3`

`E_z=int (k*z*lambda R d(theta))/r^3`

Now substitute for `r` and integrate theta from `[0,2pi]` .

`E_z=int_0^(2pi) (k*z*lambda R d(theta))/(sqrt(z^2+R^2))^3`

`E_z=(k*z*lambda R)/(sqrt(z^2+R^2))^3 int_0^(2pi) d(theta) `

`E_z=(2pi k*z*lambda*R )/(sqrt(z^2+R^2))^3`

Therefore the electric field at point `P` is entirely in the `z` -direction with magnitude:


See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team