- #1
Logical Dog
- 362
- 97
I have a very weak understanding of the particle modelling of electrical phenomena.1:[tex]
Electric field =\frac{Force}{Magnitude of charge} [/tex]
1 is The definition of electric field on a particle (?)
2:[tex] Force = \frac{Q1*Q2*K}{R^2}\\[/tex]
2 is Coulombs law, to find force between two particles.
3:[tex] {\frac{(Q1*Q2*K)}{R^2}} / (Q2) = \frac{Q1*k}{R^2}[/tex]
3 is the substitution of 2 onto 1s numerator, Force
4:[tex] \frac{Q1*k}{R^2} * D = Work done[/tex]
4 is the formula for work done. D here meaning distance, or should one use S for displacement?
5:[tex] [\frac{Q1*k}{R^2} * D] /Q2 = Voltage [/tex]
5 is work done divided by magnitude of charge, which is voltage.
My question is, since, in 5, this:
[tex] [\frac{Q1*k}{R^2}] [/tex]
Is the electric field on a particle,
it fair to say, that the voltage is the electric field on a charge * distance moved as a result of fields force divided by the magnitude of that charge?
Electric field =\frac{Force}{Magnitude of charge} [/tex]
1 is The definition of electric field on a particle (?)
2:[tex] Force = \frac{Q1*Q2*K}{R^2}\\[/tex]
2 is Coulombs law, to find force between two particles.
3:[tex] {\frac{(Q1*Q2*K)}{R^2}} / (Q2) = \frac{Q1*k}{R^2}[/tex]
3 is the substitution of 2 onto 1s numerator, Force
4:[tex] \frac{Q1*k}{R^2} * D = Work done[/tex]
4 is the formula for work done. D here meaning distance, or should one use S for displacement?
5:[tex] [\frac{Q1*k}{R^2} * D] /Q2 = Voltage [/tex]
5 is work done divided by magnitude of charge, which is voltage.
My question is, since, in 5, this:
[tex] [\frac{Q1*k}{R^2}] [/tex]
Is the electric field on a particle,
it fair to say, that the voltage is the electric field on a charge * distance moved as a result of fields force divided by the magnitude of that charge?