1. The problem statement, all variables and given/known data
a certain parrallel plate capacitator has a plate separation d and a potential difference between the plates of V. If the electric field between the plates is uniform
a) What is the strength of the electric field F
2. Relevant equations
Coloumbs Law F=k(q1q2/d^2)
Electric potential=electric potential energy/amount of charge
3. The attempt at a solution
i am not sure which equation to use here with the given variables i know that the potential difference is the change in velocity (i think thats right anyway)
Doc Al
Aug27-08, 06:18 AM
i am not sure which equation to use here with the given variables i know that the potential difference is the change in velocity (i think thats right anyway)
No, that's not right. Find the relationship between voltage, field, and distance: Work and Voltage: Constant Electric Field (http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elewor.html#c1)
tiny-tim
Aug27-08, 06:27 AM
Hi annjolino! :smile:
(if you'd spelt capacitor right, you'd have got the autolink to the PF Library … :rolleyes:)
a certain parrallel plate capacitator has a plate separation d and a potential difference between the plates of V. If the electric field between the plates is uniform
a) What is the strength of the electric field F
…
i know that the potential difference is the change in velocity (i think thats right anyway)
Nooo …
Voltage = potential difference is energy/charge, so it's the same as work done per charge … and electric field is force/charge …
and work done = force times … what? :smile:
annjolino
Aug27-08, 08:11 AM
Hmmm bad day with spelling i guess....:rolleyes:
work done = force x distance
tiny-tim
Aug27-08, 08:39 AM
Hmmm bad day with spelling i guess....:rolleyes:
… what we call … incapacitated! :biggrin:
work done = force x distance
Yes … and so F = … ? :smile:
annjolino
Aug27-08, 06:09 PM
does the F in this question equal the force of the electric field or does the F mean the E int the equation Electric field = force/charge.... i know that might osund a bit weird but the way i read the question F = E= f/c.....
i think i am going the long way about this but so far i have arranged the three formulas to read this
V=W/q which looks like V= (F/d)/(F/E)
am i headed in the right direction???
annjolino
Aug27-08, 06:15 PM
i just noticed a little hint in the question 'NB 1Vm-1 is equivalent to 1NC-1
so from the Vm-1 does that mean that the F=V/d?????
could it be that simple
tiny-tim
Aug27-08, 06:20 PM
does the F in this question equal the force of the electric field or does the F mean the E int the equation Electric field = force/charge.... i know that might osund a bit weird but the way i read the question F = E= f/c.....
Hi annjolino! :smile:
Yes … the question is very clear …
a) What is the strength of the electric field F
That means that F is the electric field. :smile:
V=W/q which looks like V= (F/d)/(F/E)
am i headed in the right direction???
Not following that. :confused:
(and what's E anyway? there's no E in the problem :rolleyes:)
Write it out clearly … like this …
V = Voltage = work done per charge
Work = force x distance
F = Electric field = force per charge
So V/F = work / force = distance = d. :smile:
annjolino
Aug27-08, 09:18 PM
part c of the question is, if the acceleration of a charged particle between the plates is a when it is halfway between the plates, why will the particle acceleration experience the same acceleration if it is only one quarter the distance between the plates? defend your answer...
Ok so i think the the particle will experience the same acceleration at any point across the field due to the fact that the electric field between the plates is uniform... as stated by the question... i am just not sure how to defend my answer... i guess i have to use an equation of sorts??
Defennder
Aug28-08, 05:18 AM
You've already answered the question. Why do you need to defend it? If you want to put it more explicitly, you just need to show how force is related to E-field and acceleration of charged particle.
tiny-tim
Aug28-08, 05:22 AM
Ok so i think the the particle will experience the same acceleration at any point across the field due to the fact that the electric field between the plates is uniform... as stated by the question... i am just not sure how to defend my answer... i guess i have to use an equation of sorts??
Hi annjolino! :smile:
Yes, that's fine … :smile:
if you really want an equation, how about good ol' Netwon's second law … F = ma … combined with the equation for force from an electric field? :wink:
ooh … EDIT:
You've already answered the question. Why do you need to defend it? …