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# 2.1: Introduction to Electrostatic Potentials

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

Imagine that some region of space, such as the room you are sitting in, is permeated by an electric field. (Perhaps there are all sorts of electrically charged bodies outside the room.) If you place a small positive test charge somewhere in the room, it will experience a force $$\textbf{F} = Q\textbf{E}$$. If you try to move the charge from point $$A$$ to point $$B$$ against the direction of the electric field, you will have to do work. If work is required to move a positive charge from point $$A$$ to point $$B$$, there is said to be an electrical potential difference between $$A$$ and $$B$$, with point $$A$$ being at the lower potential. If one joule of work is required to move one coulomb of charge from $$A$$ to $$B$$, the potential difference between $$A$$ and $$B$$ is one volt ($$\text{V}$$).

The dimensions of potential difference are $$\text{ML}^2 \text{T}^{ −2}\text{Q}^{ −1}$$.

All we have done so far is to define the potential difference between two points. We cannot define “the” potential at a point unless we arbitrarily assign some reference point as having a defined potential. It is not always necessary to do this, since we are often interested only in the potential differences between points, but in many circumstances it is customary to define the potential to be zero at an infinite distance from any charges of interest. We can then say what “the” potential is at some nearby point. Potential and potential difference are scalar quantities.

Suppose we have an electric field $$E$$ in the positive $$x$$-direction (towards the right). This means that potential is decreasing to the right. You would have to do work to move a positive test charge $$Q$$ to the left, so that potential is increasing towards the left. The force on $$Q$$ is $$QE$$, so the work you would have to do to move it a distance $$dx$$ to the right is $$−QE\, dx$$, but by definition this is also equal to $$Q\, dV$$, where $$dV$$ is the potential difference between $$x \text{ and }x + dx$$.

Therefore

$E =-\dfrac{ dV}{dx}. \label{2.1.1}$

In a more general three-dimensional situation, this is written

$\textbf{E} = -\textbf{grad}\,V=-\nabla V = - \left ( \textbf{i}\dfrac{∂V}{∂x} + \textbf{j}\dfrac{∂V}{∂x}+\textbf{k}\dfrac{∂V}{∂x} \right ) . \label{2.1.2}$

We see that, as an alternative to expressing electric field strength in newtons per coulomb, we can equally well express it in volts per meter ($$\text{V}\, \text{m}^{−1}$$ ).

The inverse of Equation \ref{2.1.1} is, of course,

$V = -\int E\,dx + \text{ constant}\label{2.1.3}$