Relationship between electric field intensity and potential

[kihr]

I find that there are two ways in which the relationship between electric field intensity and potential are expressed as mentioned hereunder:

(A) E = - dV / dx (V = potential and x is the direction along which V varies)

(B) E = - Delta V / Delta x

My understanding of the application of (A) and (B) is as follows:

(A) is the general case where the variation of V with x could be linear or non-linear, i.e. E could be constant or variable along the x direction. E represents the limiting value of an incremental value of V divided by an incremental value of x (at a given value of x) as the increment in x is made infinitesimally small. Thus E is the gradient of the V versus x graph at any given value of x.

(B) is a specific case of (A) when V varies linearly with x, when E is constant along
the x direction. In this case the value of E would be the same irrespective of the
values chosen for Delta V or Delta x.



I would appreciate if the above understanding could be commented upon / ratified. Thanks.

 

[Doc Al]

Looks like you understand things to me.

 

[kihr]

I am sorry if I have offended you in any way by my query. I genuinely request for a confirmation that my understanding is correct. Kindly advise. Thanks.

 

[cepheid]

I am sorry if I have offended you in any way by my query. I genuinely request for a confirmation that my understanding is correct. Kindly advise. Thanks.

I doubt very much that you offended Doc Al. My interpretation of his post was that he was merely stating in a very succinct way that your original post looked correct and didn't contain any obvious errors. If it did, I'm almost certain he would have pointed them out. Perhaps with the following edits the meaning (as I understood it) becomes clearer: "[It] [l]ooks like you understand things [just fine] to me."

The only thing I would add to your post is that the most general statement of the relationship between the electric field and the electric potential is:

\mathbf{E}=-\nabla V​

The other two equations are merely special cases of it, the second being common in introductory physics when students are not expected to have a grasp of vector calculus (or perhaps calculus of any kind)

 

[clustro]

Your analysis of things is correct, kihr.

 

[kihr]

Thanks very much. Looks like I had goofed up with my interpretation of the English language!

 

[cabraham]

I doubt very much that you offended Doc Al. My interpretation of his post was that he was merely stating in a very succinct way that your original post looked correct and didn't contain any obvious errors. If it did, I'm almost certain he would have pointed them out. Perhaps with the following edits the meaning (as I understood it) becomes clearer: "[It] [l]ooks like you understand things [just fine] to me."

The only thing I would add to your post is that the most general statement of the relationship between the electric field and the electric potential is:

\mathbf{E}=-\nabla V​

The other two equations are merely special cases of it, the second being common in introductory physics when students are not expected to have a grasp of vector calculus (or perhaps calculus of any kind)

This is NOT the most general relation. This only holds in the special case for conservative E fields. The more general expression is:

E = -grad V - dA/dt + (u X B).

A = magnetic vector potential, u = velocity, B = magnetic flux density.

Also, V = integral {E*dl} along a path.

Claude