Comparing Electric Potential and Electric Field

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SUMMARY

The discussion centers on the relationship between Electric Potential (V) and Electric Field (E), specifically the equation ΔV = -∫E ds. The user expresses confusion regarding the graphical representation of these concepts, questioning why Electric Field increases when Electric Potential decreases. It is clarified that the Electric Field is the negative gradient of Electric Potential, meaning that as a positive charge moves in an electric field, it loses potential energy. The sign of the potential difference is context-dependent, influencing how these graphs are interpreted.

PREREQUISITES
  • Understanding of Electric Potential and Electric Field concepts
  • Familiarity with calculus, specifically integration
  • Knowledge of the relationship between force, charge, and electric fields
  • Basic graphing skills for visualizing mathematical relationships
NEXT STEPS
  • Study the concept of electric field lines and their relationship to electric potential
  • Learn about the implications of the negative gradient in electric fields
  • Explore uniform electric fields and their characteristics
  • Review advanced topics in electrostatics, such as potential energy in electric fields
USEFUL FOR

Students preparing for exams in physics, educators teaching electromagnetism, and anyone seeking to deepen their understanding of electric fields and potentials.

MattyP
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Homework Statement



I guess this isn't like a homework question. I'm studying for an upcoming exam and feel pretty confident about most of the material, except for the relationship between Electric Potential and Electric Field. I guess the pertinent equation is as follows:


Homework Equations



ΔV=-∫Esds

The Attempt at a Solution



I understand the concept behind this. Electric potential is the negative integral of electric field. What confuses me is that my professor keeps graphing both in the positive y-axis. If electric potential is the NEGATIVE integral of electric field, wouldn't it always be the "opposite". So if graphing E vs distance and E = +1000 V/m at d = 20m and electric field is increasing as distance increases, wouldn't the electric field vs distance graph start at, say, some arbitrary y-value and then decrease as distance decreases?

This doesn't conceptually make sense to me anyway because of electric potential's equation. I would imagine that E would increase as V increased since the only difference in equation is that one is inverse squared and one is not.

I've derived V from E so that I would understand it mathematically as well. I guess my issue is the conceptual reasoning for it.

So to sum it up, why does E increase when V decreases

Sorry if this is redundant / been asked before. I know I'll probably be yelled at for messing something up my first time posting her, but I guess I'll learn
 
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The electric field is the force felt by a positive charge divided by the charge.
The potential is the potential energy divided by the charge.

These two are related in that the electric field is the (negative) gradient of the potential.
The negative sign is due to the definitions: when a positive charge moves along an electric field line, it has to lose potential energy.

The sign of the potential difference depends on the reference point so what the plot looks like depends entirely on context. \

why does E increase when V decreases
The relationship is not that simple.
i.e. in a uniform electric field, you may move between positions of varying potential, yet the E field does not change at all.
http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html
 

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