Physics Question

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afzal

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I was wondering if somebody could please help me with a physics question. The question is, how can I show that newton/coulomb = volt/meter by the use of dimensional analysis.

Thanks

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afzal said:
I was wondering if somebody could please help me with a physics question. The question is, how can I show that newton/coulomb = volt/meter by the use of dimensional analysis.

Thanks


(N/C) = (V/m) ---> V = J/C
(N/C) = (J/Cm) ---> J = units for work/energy = Fd = Newtons x Meters
(N/C) = (Nm/Cm) ---> meters cancel
(N/C) = (N/C) ---> I think that's right
 
use:
potential (in Volts) is energy gained (Joule) per unit charge (Coulomb)
(eqn: V=qPE).
so a V=J/C, and J=Nm, so V=Nm/C
thus a V/m=(Nm/C)/m=N/c

or:
potential (V) is the path integral (m) of the Electric field (N/C)
(which is just a fancy way of saying the eqn: V=Ed)
so you get the same thing: V=m*N/C, so V/m=N/C


will is 100% right. I'm just echoing him 'cause I'm trying to get to 25 posts so I can see that magic hidden forum.
 
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I could be very wrong, but here is what I think:

Newton = Force = E * Q (electric field * total charge)
Voltage = E * d (distance)

So, E * Q/ C = E * D / meters

simplify: Q/C = D / meters

both equal time.

I hope this is right, but if not, someone please correct me.
 
there's several ways to get to the right place.

side note: someone pointed out to me that if you mentally organize the electrostatic eqns (like F=kQq/r2), sometimes is easier to remember the relationships (the secondary eqns). like:

E=kQ/r2 --------x by q----> F=kQq/r2
|
x by r (distance)
|
V=kQ/r ---------x by q----> PE=kQq/r

so to go from Force to Electric field, you divide by q (so E=F/q)
or to go to potential from field, you multiply by distance (so V=Ed)
or to get to energy from from field, you multiply by both q & distance (so W=qEd)
etc.

(you probably have seen that before, but I just remember it being easier to memorize all the eqns after I saw them laid out like that)

almost there...
 
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