Electric Potential - Conceptual question

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SUMMARY

The discussion centers on calculating the distance from a point charge and the charge's magnitude based on given electric field and potential values. The electric field is 500 V/m, and the electric potential is -3.00 kV, leading to a distance of 6 meters and a charge of -2 micro-Coulombs. The negative charge is confirmed through the relationship between electric potential and charge, where the negative potential indicates a negative charge. The distinction between using electric field (E) and electric potential (V) in determining charge is also clarified.

PREREQUISITES
  • Understanding of electric field and electric potential concepts
  • Familiarity with Coulomb's law and the formula E = kq/r²
  • Basic knowledge of charge polarity and its implications
  • Ability to perform calculations involving micro-Coulombs and kilovolts
NEXT STEPS
  • Study the relationship between electric field and electric potential in detail
  • Explore the implications of charge polarity on electric field direction
  • Learn about the concept of electric field lines and their representation
  • Investigate the effects of different charge configurations on electric potential
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and potentials.

EngineerHead
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Homework Statement



At a certain distance from a point charge, the magnitude of the electric field is 500 V/m and the electric potential is -3.00 kV. (a) What is the distance to the charge? (b) What is the magnitude of the charge?

Homework Equations



The Attempt at a Solution



a) 6 meters
b) -2 micro-Coulombs
*These are correct, I mention this fact cause I don't want you to have to waste your time checking so.

My question is: for part b, if you solve for Q, the magnitude of charge, from E instead of V (E = kq/r^2), what is the visual concept behind why the charge is negative? Because obviously if you solve from V, you will get a negative 2 uC.

Of course the book asks for magnitude anyway, but the answer key gives the negative sign as well. I want to understand the concept behind it.
 
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EngineerHead said:
My question is: for part b, if you solve for Q, the magnitude of charge, from E instead of V (E = kq/r^2), what is the visual concept behind why the charge is negative?
You cannot tell the sign of Q just using E, since you were not told the direction of the field.
 
Thank you, and just to verify - because of the potential given, it is a fact that the charge is negative?
 
EngineerHead said:
Thank you, and just to verify - because of the potential given, it is a fact that the charge is negative?
Yes.
 

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