Relation between E and Potential gradient.

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Discussion Overview

The discussion centers on the relationship between electric field strength (E) and electric potential (V), particularly focusing on how potential changes with distance from a charge and the implications of charge sign on these concepts. Participants explore theoretical aspects, conceptual clarifications, and potential implications in practical scenarios.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant notes that according to the theory, E = -dv/dx implies that if both E and V are positive, the potential drop should increase, but questions this due to the behavior of field strength and potential with distance from a charge.
  • Another participant emphasizes the importance of considering the sign of the charge when discussing electric potential and potential energy.
  • A different participant asserts that electric potential is defined in terms of a unit positive charge and that potential energy will depend on the charge's magnitude and sign, suggesting a rigorous approach to signs is necessary.
  • One participant agrees that as the distance from a positive charge increases, field strength decreases, which they argue leads to a decrease in potential drop, questioning the initial claim.
  • Another participant suggests that potentials can be determined without calculating fields, stating that potentials are linear and can be manipulated like other linear functions.
  • A later reply introduces an analogy of potential as a hilltop, explaining that as one approaches a positive charge, the potential increases, with the steepness representing the electric field.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between electric field strength and potential, particularly regarding how potential changes with distance from a charge. While some agree on the importance of charge sign and the linearity of potential, others challenge the initial assumptions about potential drop and field strength.

Contextual Notes

There are unresolved assumptions regarding the behavior of electric potential and field strength, particularly in relation to distance and charge sign. The discussion reflects varying interpretations of these concepts without reaching a consensus.

harjyot
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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 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
 
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You must take the sign of the charge into account.
 
UltrafastPED said:
You must take the sign of the charge into account.

Well, yes but Electric Potential is defined in terms of a unit positive charge. The Potential Energy of a charge will depend upon its magnitude and sign.
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.
 
Isn't that what I said?
 
I didn't say I disagreed. I was just amplifying and emphasising.
 
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harjyot said:
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
Yes, this is correct.

harjyot said:
and potential drop should increase
As the field strength decreases the potential drop also decreases. I am not sure why you think otherwise.


harjyot said:
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
You don't need to calculate fields to determine the potential. Potentials are linear, so you can simply add them up or shift them around like any other linear function.
 
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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