You've come here to find your potential, haven't you? Guess what? No. This bit crushed me. It will crush second-year physics major as long as it exists. There is no sugarcoating it. That being said, let's take a look at what electric potential is! # P.D - Not a Police Department *Take a look at* [[Power, Work Done and Energy#All About Potential Energy|Potential Energy]] *before starting!* Let's begin - what are we trying to define? When a particle has a charge, being stuck in an [[Electric Fields Basics (ElectroMagnetism)|electric field]] means that there'll be a force being exerted on it. Say we had two positive charges (diagram below). One of the particles would be repulsed by the other thanks to its electric field, meaning that it'd start to move. However, both particles were stationary before, meaning that we've conjured energy up from... nothing. That's not possible, so we physicists have come up with ways to uphold the fundamentals of physics. Basically, we've defined that the kinetic energy has been converted from a 'electric potential energy'. The electric field thus has to give the particle some sort of property in order for there to be a potential energy. Remember, if the charges are stationary at the beginning: $U = -W$ Let's also define a point where the potential energy of the charge is 0. To get to this point, charge must be brought in from an imaginary point where the electric field does not act - an arbitrary infinity. Using the identity gives us: $V = -\frac{W_{\infty}}{q} = \frac{U}{q}$ This means 1 volt is equal to *1 Joule per Coulomb*, or $\frac{J}{C}$. We can rewrite this to get: $U = qV$ This is pretty much why you call a voltage a potential difference, especially if not all the power in the cell is consumed at once! We can also find any change in the potential energy as: $dU = qdV$ # Motion Through an Electric Field To get a charge to move, we need to apply work to it, meaning that it's potential energy is also going to change. See: $\Delta U = q(v_{f} - v_{i})$ Using the identity with work done we can get: $W = -q(v_{f}-v_{i})$ To simplify this let's make it so that the object isn't moving at all as we start time. Using the [[Power, Work Done and Energy#Work Done - A Brief Rundown|work-energy theorem]] the equation for work becomes the equation for kinetic energy: $E_{K} = -q(v_{f}-v_{i}) = -q\Delta V$ If the work done is a consequence of an *applied* force - that meaning we're pushing it on purpose, there's another external work done expression we have to add to what we have already: $E_{K} = -q\Delta V + W_{app}$ If the particle doesn't move, we get an expression for the applied work as just being: $W_{app} = q\Delta V$ It's a special case! Everything has to be one these days... ## The Thermo Special - Conductivity Gradients The current can be given by... $\frac{dQ}{dt} = A \frac{V}{L}$ Corrupt through a conducting wire can be calculated as: $I = \sigma A \frac{V}{L}$ $L = \sigma A \frac{V}{I} = \sigma A R$ $R =\frac{I}{\sigma} \frac{L}{A}$ Where: - $L$ is the length of the wire - $A$ is the cross-sectional area of the wire - $\sigma$ is the conductivity constant Conductance is the ratio of current to voltage. Therefore: $C = \frac{I}{V} = \frac{1}{R}$ # Work Done by an Electric Charge