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negative charge on oil droplet? watch

1. explain why when an oil droplet is in a uniform field between 2 parallel plates it experiences a negative charge? please hlp - thanks
2. anyone?
3. if the lower plate was negstive then the electrical force balances the gravitational force so it stays uniform so it's negatively charged because if it was poisitve then it will move downwords
4. excellent - i didnt realise the setup was vertical. it mentioned that this experiment was odne by milikan or something. have u heard of this before and thats how u knew or ur just real godd at physics. oh yea and wots the deal with u and physics - do u do physics or just enjoy it - i remember u said once that u wish u coyuld be caklled a physicist?
5. further question:

a second but unobserved oil drop which has an equal byut opposite charge, is 0.1mm from the first drop. show that the magnitude of the force between the two drops may be neglected in the calculations in b.)

the calculation i carried out in b. are as follows.
calculate the field strength between the plates = 9.5x10^4vm^-1
calculate the magnitude of the charge on the oil drop = 3.2 x 10^-19 C

the information given for the first drop was: weight = 3 x 10^-14N. the plates are 4.0 mm apart and the p.d between them is 380v.

sorry ive made this so confusing - i can do the questions in opart b but cant do the part in bold. info for that question is that the permitivitty is 8.9 x 10^-12Fm^-1

i will also give the answer now thats in the book just to save time later:

9.2 x 10^-20N which is much less than 3 x 10^-14N

6. just to add there is a diagram that shows a positive and negative drop 0.1mm apart vertically - if that makes any differnce - and habosh that comment i made wasnt sposed to sound so harsh - i was just wundrin wot u meant. i thought that u were at uni doing something other than physics and u were saying how much u luvd physics - i dunno?
7. (Original post by ryan750)
further question:

a second but unobserved oil drop which has an equal byut opposite charge, is 0.1mm from the first drop. show that the magnitude of the force between the two drops may be neglected in the calculations in b.)

the calculation i carried out in b. are as follows.
calculate the field strength between the plates = 9.5x10^4vm^-1
calculate the magnitude of the charge on the oil drop = 3.2 x 10^-19 C

the information given for the first drop was: weight = 3 x 10^-14N. the plates are 4.0 mm apart and the p.d between them is 380v.

sorry ive made this so confusing - i can do the questions in opart b but cant do the part in bold. info for that question is that the permitivitty is 8.9 x 10^-12Fm^-1

i will also give the answer now thats in the book just to save time later:

9.2 x 10^-20N which is much less than 3 x 10^-14N

ok I guess you shoulsd use kQq/r^2
so 9x10^9 x (3.2 E-19)2/(.1E-3)2=9.22E-20 whichi is a very small force and can be ignored as you see the force is very small like 3X10^-14/9.2x10^-20=which is like the weight is 325521times this force
8. (Original post by ryan750)
excellent - i didnt realise the setup was vertical. it mentioned that this experiment was odne by milikan or something. have u heard of this before and thats how u knew or ur just real godd at physics. oh yea and wots the deal with u and physics - do u do physics or just enjoy it - i remember u said once that u wish u coyuld be caklled a physicist?
no it's horizental like the oil drop moves from left to right and the electrical force for example is upwords between the plates while the weight is downwords
I do physics EDEXCEl and i'm a big bragger when it comes to exams something like a lightening strikes my brain and I screw them up,actully I've been kicking my self and crying for 2 weeks since phy4 and phy5 exams

and I dont love physics anymore well I like it but after the exams it really has dissapointed me ,
9. thanks so much - my physics 4 went real bad too - i do aqa b. my only hope is physics 5 now.

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