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Relation between E and Potential gradient. 
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#1
Mar114, 01:57 PM

P: 42

According to the theory,
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 xaxis 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 


#2
Mar114, 06:43 PM

Sci Advisor
Thanks
PF Gold
P: 1,908

You must take the sign of the charge into account.



#3
Mar214, 11:55 AM

Sci Advisor
PF Gold
P: 11,958

In situations like this, the thing to do is to follow the signs rigorously and not to trust to intuition. The system is quite consistent. 


#4
Mar214, 03:58 PM

Sci Advisor
Thanks
PF Gold
P: 1,908

Relation between E and Potential gradient.
Isn't that what I said?



#5
Mar214, 04:19 PM

Sci Advisor
PF Gold
P: 11,958

I didn't say I disagreed. I was just amplifying and emphasising.



#6
Mar214, 05:34 PM

Mentor
P: 16,999




#7
Mar414, 04:38 PM

P: 64

Think of potential as the energy required to bring a charge from infinitely far away to the point in question. The closer you get to the point in question (say a positive charge), the higher your potential. Now think of the potential as a hilltop. At infinity it is flat, but as you get closer to your charge, the hill gets steeper. This "steepness" is the field.



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