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February 18[edit]

Stationary Charge in an Electric Field[edit]

I can't seem to understand what this physics question, from my textbook, is asking:

   What must the charge (sign and magnitude) of a 1.45 g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N/C?

I don't understand why the charge would be stationary, since its in the electric field and no other forces are operating on it. Presumably, wouldn't it continue moving towards the source of the field or away from it? OldTimeNESter (talk) 01:49, 18 February 2020 (UTC)[reply]

Gravity is still acting on it, and it has a mass of 1.45 g So the implication is that the electrical force balances this, i.e. it acts upwards and is equal to the weight. Andy Dingley (talk) 02:05, 18 February 2020 (UTC)[reply]

The use of the term "downward-directed" implies the presence of a gravitational field. Still, I think textbooks should be explicit about whether this is the standard Earth gravitational field. --Lambiam 08:46, 18 February 2020 (UTC)[reply]

It's a poorly-worded question (what's the 'magnitude' of an electric field? The appropriate term there would be intensity.) Also I can't answer the sign of the charge because I don't know what their convention of 'downward-directed' means. The usual term would be to state positive or negative uppermost or downwards. Andy Dingley (talk) 11:54, 18 February 2020 (UTC)[reply]

The question is basically asking what the charge of the particle needs to be to levitate it if its mass is 1.45 g. The simplest way to solve confusing problems is by dimensional analysis. The question is asking for charge, measured in Coulombs (C). You have a number that gives Newtons per Coulomb, which means you just need to find some Newtons and divide it by that number to get your answer. Your Newtons will come from the force of gravity, which is gotten by multiplying the mass by 9.8 m/s², the acceleration due to gravity. 1.45*9.8 = 14.21 N. 14.21 N / 650 N/C = 0.0218 C. So your answer is 0.0218 C. I believe the standard sign convention is to indicate the direction of the field as directed away from positive, so that means that + is up and - is down, so your charge would need to be negative. The answer would be -0.0218 C. Again, the best advice I can give to solving any problem of this nature is to focus on the units, and figure out what unit your answer is in, and then set up the math so the units algebraically cancel to give you the correct unit, the rest of the math should work out. --Jayron 32 13:00, 18 February 2020 (UTC)[reply]

You got your units mixed up. It would be 0.00145Kg * 9.8m/ss = 0.01421N / 650N/C = 0.0000218C. Anon — Preceding unsigned comment added by 208.74.48.252 (talk) 14:51, 18 February 2020 (UTC)[reply]

Yes, you're right. Thank you for that correction. --Jayron 32 15:14, 18 February 2020 (UTC)[reply]

Thanks, everyone, for the clarification. I didn't know that I was supposed to consider gravity as another force acting on the charge. OldTimeNESter (talk) 16:12, 18 February 2020 (UTC)[reply]

If something measures forces and masses, gravity is going to show up somewhere, usually. There's only a few kinds of force calculations you do with mass; inertia (which doesn't require gravity, but will instead have to give you an acceleration value per F=ma), falling (which does, in which case the "a" is 9.8 m/s/s. This is a variation on that) or friction (which usually does in the calculation of the normal force, which is just the falling calculation + some trigonometry). --Jayron 32 16:17, 19 February 2020 (UTC)[reply]

A helminthology term: zyboloic[edit]

What in the world does zyboloic mean? Khemehekis (talk) 11:52, 18 February 2020 (UTC)[reply]

Not helpful, I know, but I think it means the worm's prostomium has a shibboleth on it. Probably in the shape of a zybol. Given that that article has probably the highest density of simply unknown words on it of any WP article I've read, I think this is a word which is just unused outside of its narrow (and wiggly) field. A prostomium may be zygolobic too,[1] and the two terms aren't exclusive (some are both together).[2] Andy Dingley (talk) 12:07, 18 February 2020 (UTC)[reply]

I suspect that it is a strange typo for zygolobic, meaning that the lobes of the prostomium are joined (zygo- + lobe + -ic). --Lambiam 20:34, 18 February 2020 (UTC)[reply]

Introduced by Wilhelm Michaelsen [3] for describing oligochaeta, don't know if it would apply to other annelids.—eric 01:58, 19 February 2020 (UTC)[reply]

But that's zygolobic, and we're looking for zyboloic. The Korean source lists several which are both at once. Andy Dingley (talk) 02:33, 19 February 2020 (UTC)[reply]

And all the above links are to oligochaetes. For some reason i thought zyboloic was used when describing a different class.—eric 04:55, 19 February 2020 (UTC)[reply]

Someone introduced this as a misspelling, which then was copied by someone else. It is telling that GBS gives no hits. There is also no Greek root this could derive from. --Lambiam 13:41, 19 February 2020 (UTC)[reply]

I wrote an email to the creator of the website and received a reply: "Yes it is spelling error on my part. Apologies." --Lambiam 07:59, 21 February 2020 (UTC)[reply]