Why do we have a charge in the denominator of equation for voltage?

Click For Summary
The discussion centers on the presence of charge in the denominator of voltage equations, questioning its necessity since voltage and electric potential are not directly dependent on charge. It explains that electrostatic potential is defined as electrostatic potential energy per unit charge, making it easier to conceptualize energy in a scalar field. The analogy with gravitational potential illustrates that while energy depends on mass, the potential itself is independent of it. The conversation also touches on the foundational aspects of physics, suggesting that definitions may vary based on what is considered fundamental. Ultimately, the inclusion of charge in voltage equations serves as a useful framework for understanding energy interactions in electric fields.
Callmelucky
Messages
144
Reaction score
30
Homework Statement
Why do we have a charge in the denominator of equations for voltage and el. potential if both voltage and el. potential are not dependent on charge?
Relevant Equations
U=W/q, fi=Eep/q (fi=el. potential, Eep= el. pot. energy, U= voltage)
Why do we have a charge in the denominator of equations for voltage and el. potential if both voltage and el. potential are not dependent on charge?
Is it just because that was the only way to derive the formula for voltage and then we realized we don't need q? U=W/q --> U=eqd/q.
 
Physics news on Phys.org
It's a definition. Electrostatic potential ##V_e## is electrostatic potential energy ##U_e## per unit charge. The energy does depend on the charge but it is easier to think of a scalar field ##V_e## such that when we place charge ##q## at some point in space, its energy will be ##U_e=qV_e##.

You have already encountered this idea. Compare with something familiar, gravitational potential. Near the surface of the Earth it is ##V_g=gh##. When one puts mass ##m## at height ##h##, its gravitational potential energy is ##U_g=mV_g=mgh.##
 
  • Like
Likes member 731016 and Callmelucky
To add to @kuruman's reply…
It depends what you take as fundamental. If you take energy, distance and time as fundamental then you would define the mass of an object as the work needed to accelerate it to a given speed.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...