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