According to the theory,(adsbygoogle = window.adsbygoogle || []).push({});

E= -dv/dx

or E.dx = -dv

So if both are positive, the potential drop should increase.

But as we know, if a positive charge is placed, as the distance from it keeps on increasing, field strength starts decreasing and potential drop should increase But this is contradictory right?

Similarly if we make a diagram of field lines along the x-axis and place a sample charge at (a,0,0) and move it to (b,0,0) such that b>a, how can we predict which is at greater potential : a or b? The field value isn't given

**Physics Forums - The Fusion of Science and Community**

# Relation between E and Potential gradient.

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Relation between E and Potential gradient.

Loading...

**Physics Forums - The Fusion of Science and Community**