# Calculating Electric Fields: Help

## Homework Statement

Question: An electron and a proton are fixed at a separation distance of 925nm. Find the magnitude and direction of the electric field at their midpoint.

E= Fe/q

## The Attempt at a Solution

(8.988 x 10^9) x (q/(9.25 x 10^-7)2)

This is the equation i have to solve for it but I don't know what q is in this situation. Help?

## Answers and Replies

q in this case is the fundamental unit of charge. Both the electron and proton have the same magnitude of charge, but opposite. Since you're dealing with a single electron and proton, q is just 1.60E-19 Coulombs; sign depending on which particle you're examining.

But since they have equal and opposite charges, wouldn't that make q=0 which then makes the entire solution 0. Or does it have to do something with their midpoint in this problem?

These are non-moving point charges. There are a couple of other relationships that I might try to use to examine this scenario. One is the electric field of a point charge. The other is the forces between two charged particles separated by a distance r.

Try it. :-)

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But since they have equal and opposite charges, wouldn't that make q=0 which then makes the entire solution 0. Or does it have to do something with their midpoint in this problem?

q would only = 0 if the two particles were coincident. But to hopefully get you pointed in the right direction: the electric field of a dipole is not zero (except in the limit as r-->infinity.) Moreover, electric fields of point charges superimpose, just as forces do. So the pointcharge model--if you consider both the proton and the electron simultaneously--should get you there.