We know $$E = q V$$ where $E$ is the energy (in Joules), $V$ is the potential difference (in Volts), and $q$ is the charge. Why is this equation true and how we prove it?
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asked Aug 10, 2013 at 15:48
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$\begingroup$ Voltage is defined as potential energy per charge: $V=\frac{E}{q}$. There's nothing to prove. But, I think you are really looking for the gain or loss in potential energy formula: $\Delta V = \Delta E / q$. you just consider the voltage at point A, and the voltage at point B then subtract them. $V_A- V_B=\frac{E_A}{q} - \frac{E_B}{q}$, which is the same thing as: $\Delta V = \Delta E / q$. Some physics notes online are a little bit sloppy about adding the $\Delta$. using some algebra: $\Delta E = q \Delta V$ $\endgroup$– picoCommented Jan 5, 2020 at 14:47
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3 Answers
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There are various ways to decide which of the assumptions are primary and which of them are their consequences but $E=VQ$ may be most naturally interpreted as the definition of the potential.
The potential energy is a form of energy and the potential (and therefore voltage, when differences are taken) is defined as the potential energy (or potential energy difference) per unit charge, $V = E/Q$. That's equivalent to your equation. The potential energy is proportional to the charge essentially because of the linearity of Maxwell's equations (the superposition principle). Once we know about the proportionality, we must just give a name to the proportionality factor between $E$ and $Q$ and we simply call it potential (or voltage).
answered Aug 10, 2013 at 15:51
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There are several (equivalent) ways to look at this. One is to say that for any conservative force $\mathbf{F}$, one can define the potential energy $E_p$ as an associated potential field such as $\mathbf{F}=-\frac{\partial E_p}{\partial r}$, or maybe more formally $\mathbf{F}=-\nabla(E_p)$. That's no more than a definition of the potential energy. (Electrostatic forces and gravitational forces have that in common that they are conservative and an associated potential function exists).
At the same time electrostatic forces are (experimentally) observed to be $\mathbf{F} = q\mathbf{E}$.
Last definition is the electric potential field U : it is also defined as $\mathbf{E}=-\frac{\partial U}{\partial r}$, or maybe more formally $\mathbf{E}=-\nabla(U)$.
When one puts all these together, $E_p = qU$.
I am not sure if this is a proof though, but maybe more the consequence of various definitions of useful quantities and concepts in physics.
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electric field strength is
$$E=\frac Fq=\frac Vd$$
with $V$=voltage, $d$=distance between charged plates
\begin{align}
\frac Fq&=\frac Vd \\
Fd&=qV
\end{align}
but $Fd$=energy
$$\therefore {\rm energy}=qV$$
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5$\begingroup$ The question isn't about capacitors, so we don't necessarily have any charged plates. Manipulating equations like this doesn't really constitute a proof of anything unless you decide what you consider to be your fundamental assumptions. $\endgroup$– user4552Commented Oct 17, 2014 at 23:03