Question

How can qnAv_d delta t equal delta Q?

Answer 1

\[\Delta Q=qnAv_d\Delta t\]

q is in coulombs
n is just a number
\(\Delta t=time\)
A=\(m^2\)
\(v_d=\frac m s\)
that adds up to \(m^3C\) and not C

Answer 2

@UnkleRhaukus

Answer 3

\(m^3\) is volume so \(v_dtA=volume\)

Answer 4

volume times q?

Answer 5

where'd you get that original equation from?

Answer 6

\[I=\frac{\Delta Q}{\Delta t}=qnAv_d\]
relationship b/w current and drift speed

Answer 7

i think \(n\) has dimensions of \([\text{electrons}/\text m^3]\)

Answer 8

oh yes! So it's not dimensionless?

Answer 9

oh it says in my book that n is the number density

Answer 10

electrons per cubic meters....makes sense

Answer 11

i guess that makes the current \(I\) have dimensions of \([\text{electron}\cdot\text {Amps]}\). which also makes sense

Answer 12

yep.....current density is \(J\) is current per area so \(\frac I A =\frac{qnAv_d}{m^3}=\frac{Q/t}{m^3}=\frac C{t\cdot m^3}\) does this make sense? or should the t not be there?

Answer 13

@UnkleRhaukus

Answer 14

\[t[\text s]\]

Answer 15

\[I[\text A]=I[\text {C/s}]\]

Answer 16

oh that's right! durrrr.....sorry

Answer 17

hold on

Answer 18

yes, agreed.

Answer 19

\[J=\frac IA\left[\frac{\text A}{\text m^2}\right]\]

Answer 20

my mistake....m^2 not m^3...yep got it

Answer 21

all good now?

Answer 22

yes sir. Thank you Sir Rosser :)