How Does Electrostatic Induction Affect Electric Fields Near Conductors?

[narfarnst]

Okay, I have a feeling this is a pretty easy problem to solve, the only trouble is, my professor worded it really weirdly, so I only have half an idea of what it's asking.
Any ideas?

-----------
Let the Electric field be E at a point P near a conductiong material. Because of electrostatic induction, the force on a charge q at P will not yield E. Show, however, that the average of the fields of two equal and opposite charges +q and -q gives the correct value: E=(F_q+-F_q-)/2q.
------------
From what I can gather, it's asking me to prove the average E field of a dipole is that.

Thanks.

EDIT:
Nevermind, I just figured it out.

 

[jbsteele]

To prove this, we can use the formula for electric field of a point charge, E = kq/r^2, where k is a constant, q is the charge, and r is the distance.By summing the electric field of the +q and -q charges, we get: E = (kq/+q)/r^2 + (kq/-q)/r^2Simplifying, we get:E = 2k/r^2Thus, the average electric field of a dipole is E = 2k/r^2, which is equal to the given equation of E=(Fq+-Fq-)/2q.