Potential difference between plates

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Homework Help Overview

The discussion revolves around a problem involving a parallel plate capacitor, specifically focusing on the relationship between electric field strength, potential difference, and plate separation. Participants explore the implications of uniform electric fields and the relevant equations that govern these concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the appropriate equations to relate electric field strength, potential difference, and distance. There is uncertainty regarding the definitions of variables and the correct application of formulas. Some participants question the interpretation of the variables involved, particularly the meaning of 'F' in the context of the electric field.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants have offered clarifications on the relationships between voltage, electric field, and work done, while others are still seeking to understand the implications of the uniform electric field on particle acceleration.

Contextual Notes

Participants note spelling errors and misunderstandings of terminology, which may affect clarity. There is also a hint in the problem regarding the relationship between units of electric field and force, prompting further exploration of these concepts.

annjolino
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Homework Statement


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


Homework Equations



Coloumbs Law F=k(q1q2/d^2)
Electric potential=electric potential energy/amount of charge


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 that's right anyway)
 
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you say capacitator, I say capacitahtor … !

Hi annjolino! :smile:

(if you'd spelt capacitor right, you'd have got the autolink to the PF Library … :rolleyes:)
annjolino said:
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 that's 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:
 
Hmmm bad day with spelling i guess...:rolleyes:

work done = force x distance
 
annjolino said:
Hmmm bad day with spelling i guess...:rolleyes:

… what we call … incapacitated! :biggrin:​
work done = force x distance

Yes … and so F = … ? :smile:
 
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?
 
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
 
annjolino said:
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 …
annjolino said:
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:
 
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??
 
  • #10
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.
 
  • #11
… you have a capacity for sagacity …

annjolino said:
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:
Defennder said:
You've already answered the question. Why do you need to defend it? …

Hi Defennder! :smile:

hmm … let's see … :wink:

i] 'cos the question says so … :rolleyes:

ii] against wild animals … :eek:

iii] 'cos it can't defend itself … :blushing:

iv] it's what Cicero would have done … :approve:

v] look who's talking … Defennder :biggrin:​
 
Last edited:

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