This is an old revision of the document!
Example: Electric Field from a Negative Point Charge
Suppose we have a negative charge −Q. What is the magnitude of the electric field at a point P, which is a distance R from the charge? Draw the electric field vector on a diagram to show the direction of the electric field at P.
Facts
- There is a negative charge −Q.
- The point P is a distance R away from the charge, with the orientation shown below in our representation.
Lacking
- Magnitude of the electric field at P.
- Direction of the electric field at P.
Representations
- The electric field from the point charge can be written as →E=14πϵ0qr2ˆr,where q represents our charge (−Q in this case), r is our distance (R), and ˆr is the unit vector pointing from the point charge to P.
- We can represent the electric field in our diagram with an arrow, since the electric field at a specific point is a vector.
Solution
The electric field at P is given by: →E=14πϵ0qr2ˆr
We can plug in our charge (−Q) and the magnitude of the separation vector (R) to get: →E=14πϵ0(−Q)R2ˆr
This leaves us to find the direction of ˆr. The first thing to do would be to draw in the vector →r−Q→P. This vector points from the charge -Q to Point P since P is where we want to find the electric field. Now we need to define a coordinate axis. We could pick the normal x- and y-axes, but this would make writing the →r and ˆr more difficult because there would be both x and y components to the r-vector (since it points in some diagonal direction).
Instead, let's pick a coordinate direction that falls along the same axis as the →r−Q→P. Since this is a coordinate direction that we're naming, let's call this the ˆs direction. That means that →r−Q→P points in the ˆs direction, so ˆr=ˆs. Plugging this into our electric field equation gives: →E=14πϵ0−QR2ˆs
This gives the magnitude of the electric field as |→E|=14πϵ0QR2
