Does a Voltmeter Measure the Potential Difference Between Two Points?

[Niles]

Homework Statement

Hi all.

Lets say a voltmeter registers (measures) the following quantity:

$$V = \int_a^b{\textbf E\cdot d\textbf l}$$

Does this mean that it measures V = V(b)-V(a), i.e. the potential at b subtracted by the potential at a?

Thanks in advance.Niles.

 

[gabbagabbahey]

The conventional definition of voltage is:

$$V(b)-V(a) = -\int_a^b{\textbf E\cdot d\textbf l}$$

Which means your voltmeter would actually measure -(V(b)-V(a))=V(a)-V(b)

 

[Niles]

Hmm, isn't this only for two conductors? I mean, let's say a resistor is placed in between the points 'a' and 'b' on a wire. The voltmeter measures V as given in my original post, and the current runs from point 'b' to 'a'. Then sure V > 0, and thus you definition can't work?

Thanks for replying.

 

[gabbagabbahey]

If the current runs from 'b' to 'a', then two things must be true:

(1)The potential at 'a' is lower that the potential at 'b', and thus V(b)-V(a)>0

(2)The component of E parallel to the wire must point from 'b' towards 'a', that means $$\int_b^a{\textbf E\cdot d\textbf l}>0$$, so $$\int_a^b{\textbf E\cdot d\textbf l}<0$$

Therefore, $$V(b)-V(a)=-\int_a^b{\textbf E\cdot d\textbf l}>0$$

See for example, Griffith's 'Introduction to Electrodynamics 3rd ed.' eq. 2.22

 

[Niles]

So when I have an integral like:

$$V(b)-V(a)=-\int_a^b{\textbf E\cdot d\textbf l},$$

then it means that 'dl' points from lower limit to upper limit? And this is always true?

 

[gabbagabbahey]

So when I have an integral like:

$$V(b)-V(a)=-\int_a^b{\textbf E\cdot d\textbf l},$$

then it means that 'dl' points from lower limit to upper limit?

Yes, if you go from 'a' to 'b' along some path, then dl points tangent to that path in the general direction of 'a' to 'b'. (remember, the path (or wire) can be curved every which way, so it's inaccurate to say that at every point along the path dl points from 'a' to 'b', but certainly for a straight path (or wire), dl will point from 'a' to 'b')

 

[Niles]

Thanks!