Voltage

In a topological map, you could probably guess that the steepest path downwards/upwards is perpendicular to the lines.

The constant voltage lines works in a similar way.

If we have an object in the air

Object

_{Air}

..ground..

What is that object's gravitation potential energy?

… … …

You will realize that I just asked a very dumb question. This is because that energy must be relative to something. You must raise the object (giving us the \(\Delta h\) part of \(gpe = mg\Delta h\) to gain gravitational potential energy.

\definition{Electrical Potential}{\(\frac{V}{C}\)} Voltage is a measure of how much electric potential energy (yes, it is an energy (in \(J\), joules)), would change per Couloub of energy that is moved through.

Recall the energy example above. When you raise an object of \(1kg\) from a place with elevation (\(\Delta h\)) \(10m\) to \(100m\), you could represent the change in gravitation potential energy of that operation as \(mg \Delta h_1 - mg \Delta h_0 = m(g \Delta h_1-g \Delta h_0) = 1kg(9.8\frac{m}{s^2} 100m - 9.8\frac{m}{s^2} 10m)\). Where, \(g\Delta h_1\) is a unit \(\frac{m^2}{s^2} = \frac{J}{kg}\). Proving this last relationship is left as an excercise to the reader.

Funny way to write it, I know. But, we could take the equation \(m(g \Delta h_1-g \Delta h_0)\) and use it as a perfect analogy for using electric potential.

The amount of electric potential energy that would change by moving an object of charge \(1C\) from a place with voltage (\(\Delta V\)) \(10V\) to \(100V\), is \(Q_2(V_1-V_0) = 1C (100V-10V)\), where Voltage, \(V\), represent the energy potential — analogous to, drumroll please, \(\frac{J}{kg}\), except this time its \(\frac{J}{C}\).

\definition[Where $Q_2$ is the the charge in Coulombs $C$ of the test charge, and $V_1$ and $V_0$ are the electric potential values of the points the charge is being moved to and from]{Electric Potential Energy}{$Q_2(V_1-V_0)$} \definition{Electric Potential, Volts}{\(V = \frac{J}{C}\)}