Electric field question

1. Oct 25, 2012

Von Neumann

1. Electric field caused by a charge (or group of charges) is defined in terms of its effect on a test charge, given by the formula: E ⃗=lim┬(q_(0→0) )⁡〖F ⃗/q_0 〗

I understand that the point charge must be infinitesimally small in order not to affect the field in question, so it makes sense to take the limit of q$_{0}$ to 0. However, I do not understand why my book states the following: "If the charge(s) creating the field could be considered fixed in place, then we would not have to assume the test charge is small." Can someone please clarify?

Last edited: Oct 25, 2012
2. Oct 25, 2012

SammyS

Staff Emeritus
Hello Von Neumann. Welcome to PF !

It's not that the "test charge", q0, changes the field by producing a field of its own, it's that the test charge may alter the field produced by the other charges, if it, q0, alters the configuration of the other charges. That's particularly a problem if the field is produced by charges on a conductor.

3. Oct 27, 2012

Von Neumann

Thanks for the reply, and I'm glad to be a part of PF!

I'm still confused on this, though. You say that the test charge, q0, does not change the field by the production of its own electric field, and that the problem to consider is the test charge altering the configuration of other charges. However, how is it possible for the test charge to alter the configuration of other charges without an electric field of its own to exert a force in order to cause such an alteration? Additionally, going back to my original question, why is the charge not required to be infinitesimal for the case when the charges are stationary?

4. Oct 27, 2012

SammyS

Staff Emeritus
I suppose I could have made my statement clearer.

The purpose of the test charge is to detect whatever electric field was present prior to the test charge being introduced into the situation. You're not interested in the field produced by the test charge.

Suppose you place a test charge, q0, into a pre-existing static electric field, E. The force, F, exerted on the test charge by the pre-existing static electric field, is F = q0E, and that has no dependence upon any electric field produced by the test charge. Yes, placing the test charge at a location will alter the electric field, but the purpose of the test charge is to measure the field which existed prior to introducing the test charge. If the locations of the charges producing the pre-existing electric field are unaltered by the introduction of the test charge, then F/q0 gives an accurate measure of pre-existing electric field at the location of the test charge.

However, if the charges producing the pre-existing electric field can have their locations altered by the field produced by introducing the test charge, then you want to use as small a test charge as is practical, or better yet, if you can do it, use a sequence of test charges to determine $\displaystyle \lim_{q_0\,\to\,0\,}\, \frac{\textbf{F}}{q_0}\ .$

5. Oct 29, 2012

Von Neumann

So basically, in the case that you are calculating the pre-existing electric field produced by static charges, the test charge does not alter the configuration and therefore does not need to be infinitesimal. Additionally, in the case that you are calculating the electric produced by moving charges, the test charge does alter the configuration and does need to be infinitesimal. Is this right? I'm not sure whether an altering of the configuration of charges depends on the magnitude of the pre-existing electric field or a velocity associated with the charges. Why not always use a sequence of test charges to be sure that the configuration of charges does not alter?