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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

715.0. "conceptual electric field question" by EAGLE1::BEST (R D Best, Systems architecture, I/O) Fri Jun 05 1987 20:19

  In Ralph Morrison's book 'Grounding and Shielding in Electronic
Instrumentation', 3rd edition, the author makes the following
statement about electric fields (I paraphrase since I don't have it
in front of me):

	In order to measure an electric field, a test charge must be
	used.  The test charge must be kept small to avoid disturbing
	the field that is to be measured.

  I don't understand this.  My understanding of how electric fields
work is as follows:


	If you want to calculate the force on a particular charge,
	add up (or integrate) all the electric field contributions
	from all the OTHER charges in the system at the point where the
	particular charge is.  This electric field is then multiplied
	by the charge of the particular charge and this gives you the
	force (or equivalently a measure of the field).

  Morrison's statement seems to imply that somehow the value of the
particular charge takes part in determining the force on itself, but
this seems implausible.  Is there some subtlety that slipped by me
in Emag or is the author mistaken or just being unclear ?

			/R Best

T.R	Title	User	Personal Name	Date	Lines
715.1		BANDIT::MARSHALL	hunting the snark	`Fri Jun 05 1987 22:23`	8
	Electric fields can't act on a chargeless particle. I.E. a chargeless particle would not experiance any force. / ( ___ ) /// /
715.2	RE: .0 -- You are both correct	CAADC::MARSH	Jeffrey Marsh, DTN 474-5739	`Sat Jun 06 1987 22:29`	6
	I think the root of the confusion here is that Morrison is talking about measuring the field, and you are talking about calculating it. If you wish to measure the field, it is important that the test charge be kept small because it can induce charges and polarizations that change the original field. Your statement about calculating the field is correct.
715.3	I think I see; did I get this right ?	EAGLE1::BEST	R D Best, Systems architecture, I/O	`Tue Jun 09 1987 14:41`	30
	re .1, .2: Aha. I think I see. Let me see if I can restate in my own words. It is the total field (i.e. that due to ALL charges actually present) that is subject to Gauss law. Conceptually, we can measure the field at a particular point by calculating the E field at that point due to all other charges. However, if a test charge were actually at that point, the total E field would be different due to the effect of the test charge. The presence of the test charge would cause other free charge to move around until Gauss law was satisfied anew. But this new charge and field distribution would be different from that when the test charge was absent. For example, (if I understand this correctly) physics books always state that the charge distribution on a conducting sphere is uniform and that the force on a point charge at a distance r from the sphere is therefore Q1Q2/(4piepsilonr^2). It seems that this calculated force isn't quite right because the presence of the point charge would modify the charge distribution on the conducting sphere and therefore that the field due solely to the charge distribution on the sphere (i.e. the field that causes the force on the point charge) is DIFFERENT from that of a uniform distribution. Is this a correct interpretation ? If so, this IS a subtlety that had escaped me. This looks like an example of "you can't observe something without disturbing it".
715.4	It's all in the limit...	TSG::BRADY	Bob Brady, TSG, LMO4-1/K4, 296-5396	`Tue Jun 09 1987 18:55`	9
	Technically the field is the derivative of the force experienced with respect to the particle's charge, E = dF/dq, thus the limit of F on q at a given point as q->0. Any q ^= 0 of course experiences F ^= 0 and disturbs the field as well; as q -> 0 this disturbance, and F, -> 0 as well, but the ratio of F/q approaches some nonzero limit we call the "field strength". Same problem with gravity, any other field...
715.5	this is probably true for fluids too	EAGLE1::BEST	R D Best, Systems architecture, I/O	`Tue Jun 09 1987 20:14`	27
	re .4: > Technically the field is the derivative of the force experienced > with respect to the particle's charge, E = dF/dq, thus the limit of > F on q at a given point as q->0. Any q ^= 0 of course experiences F ^= 0 > and disturbs the field as well; as q -> 0 this disturbance, and F, -> 0 > as well, but the ratio of F/q approaches some nonzero limit we call the > "field strength". > Same problem with gravity, any other field... With gravity, it's somewhat different, I think (at least for solids). Because solids are generally treated as rigid bodies, the distribution of mass doesn't shift in response to an external gravitational field, or at least the effect is generally neglected. So, this behavior (i.e. mass or charge moving around to a new equilibrium configuration) would probably not be considered in a first order analysis of gravity effects. On the other hand, for fluids, the situation seems more analogous to that of electric charge. For example, ocean water on the earth's surface can shift in response to the gravitational forces of the moon and other planets. It would seem then that to calculate tidal effects such (fluid) mass shifting would have to be accounted for. I'd bet that the computation of the resulting charge or fluid distributions isn't easy.