# Compute the electric field in air

1. Jul 29, 2005

### in10sivkid

hi I don't exactly know how to approach this problem

Compute the electric field in air midway bewteen two point charges of 20*10^-8C and -5*10^-8C seperated by a distance of 10cm

i was thinking of finding V of both the charges and finding the difference between them

so V1 = k(20*10^-8C)/(.1m) - k(-5*-10^-8C)/(.1m)

where k = 9.0*10^9 N*m^2/C^2

then you know V then you can do V/r = E

am i even close???

thanks :)

2. Jul 29, 2005

### LeonhardEuler

The $\vec{E}$ field at a point due to multiple charges is the sum of the $\vec{E}$ fields from each charge. You do not have to worry about the voltage. As far as the equation V/r=E, I have never heard of it and I am pretty sure that it can only be valid for point charges or spheres, because otherwise I don't even know what "r" refers to.

Last edited: Jul 29, 2005
3. Jul 29, 2005

### in10sivkid

ok then...what formula do i use to calculate E fields given this information...thats how i was thinking to get to E using that equation

electric field = voltage/radius = E = V/r by the way
i got it from my professor in lecture

4. Jul 29, 2005

### OlderDan

In general the relationship between electric potential and electric field is that the change in potential is the work per unit charge to move a charge through an electric field. The customary choice for potential of a point charge is to say the potential is zero infinitely far from the charge. When that choice is made, your equation E = V/r is valid for a single point charge. In your problem, the potential at a point would be the sum of the two potentials from the two charges. If those charges were equal and opposite (they are not in your problem) the potential would be zero midway between them, but the electric field would not be zero. You cannot use E = V/r except for one point charge or a spherically symmetric charge distribution.

What you want to do in your problem is work directly with the electric fields. The electric field from a point charge is the force per unit charge on a test charge used to sample the field. From Coulomb's law this is a vector equal to the Coulomb force on the test charge divided by the test charge. The magnitude is kQ/(r^2) and points toward the charge for negative charges and away from the charge for positive charges. You must add the two vector fields to find the toal field created by the two charges. This is not difficult at the point midway between the two because the field directions are parallel.

5. Jul 29, 2005

### in10sivkid

ok so would this be good then

i did

(9*10^9 N*M2/C2)(20*10-8 C)/(.05)^2 - (9.0*10^9 N*M2/C2)(-5*10^-8 C)/(.05)^2???