Electric potential for a point charge

We previously obtained the electric potential energy of a system of two point charges from the work needed to move the charges from being infinitely far away and assemble them into a configuration \(r\) distance apart: \[\begin{equation} U_E = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r} = \dfrac{k q_1 q_2}{r}. \end{equation}\]

If we divide by the test charge, we obtain an expression for the electric potential of a single point charge, \(q\): \[\begin{equation} V = \dfrac{q}{4\pi\epsilon_0 r} \end{equation}\] where \(\epsilon_0\) is the permittivity of free space and \(r\) is the distance from the charge. When written in this way, the electric potential represents the work per (test) charge it would take to move a test charge from infinitely far away to a distance \(r\) from the original charge.

The electric potential difference between two points is the change in electric potential energy per unit charge when a test charge is moved between the two pointsEquation 11 on the official AP Physics C Electricity & Magentism equation sheet

: \[\begin{equation} \Delta V = \dfrac{\Delta U_E}{q}. \end{equation}\]

Electric potential difference can result from chemical processes that cause positive and negative charges to separate, such as in a battery, due to corrosion of a metal, or as the result of biological pumping of ions; or it can occur from physical processes like the transfer of charges, photoelectric effects, physical stripping of electrons, etc.

Superposition applies!

The electric potential due to multiple point charges can be determined using scalar superpositionEquation 8 on the official AP Physics C Electricity & Magnetism equation sheet

of the electric potential due to each of the point charges, e.g. for a collection of \(i\) charges, \[\begin{equation} V = \dfrac{1}{4\pi\epsilon_0} \Sigma_i \dfrac{q_i}{r_i} \end{equation}\]

Similarly, expressions for the electric potential of charge distributions can be found using integration and the principle of superposition: \[\begin{equation} V = \dfrac{1}{4\pi\epsilon_0} \int \dfrac{dq}{r}. \end{equation}\] This form is useful when you have some distribution of charge per unit length (\(\lambda, \unit{\coulomb\per\meter}\)), charge per unit area (\(\sigma, \unit{\coulomb\per\meter\squared}\)), or charge per unit volume (\(\rho, \unit{\coulomb\per\meter\cubed}\))

⊕AP Physics C Electricity & Magnetism only expects students to use calculus to find the electric potential resulting from the following charge distributions and locations: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis; a thin ring of charge at a location along the axis of the ring; a semicircular arc or part of a semicircular arc at its center; and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector.

Relationships between \(\vec{E}\) and \(\phi\) (\(\phi=V\)) in 3D (optional)

⊕Not testable on the AP Physics C Electricity & Magnetism exam but really useful.

More generally, we might write the potential as a scalar function of multiple variables, \(\phi=\phi(x,y,z)\). In multiple dimensions, eq. 2 is written as \[\begin{equation} \vec{E} = - \vec{\nabla} \phi \end{equation}\] where the \(\vec{\nabla}\) symbol is spoken as “del” and is given by \(\vec{\nabla}=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\) in Cartesian coordinates. It is a vector operator, and in this case results in the gradient of the potential (also sometimes written as \(\text{grad} \phi\). This means that electric field lines point down the gradient of potential.

Going the other way, from electric field to potential, is still the same: \[\begin{equation} \phi = - \int \vec{E} \cdot d\vec{s}. \end{equation}\] but in this case we can see the dot product in action as the path we take may or may not be parallel to the electric field.

There’s one other really cool use of this. Recall Gauss’s Law, which we know as \(\oint \vec{E}\cdot d\vec{a} = 4\pi k q_{enc}\), in differential form is: \[\begin{equation} \vec{\nabla}\cdot \vec{E} = 4\pi k\rho = \dfrac{\rho}{\epsilon_0}, \end{equation}\] in other words, the divergence of the electric field (\(\text{div} \vec{E}\)) is proportional to the volumetric charge density \(\rho\).

Since we are also writing \(\vec{E}=-\vec{\nabla}\phi\), \[\begin{equation} \vec{\nabla}\cdot \vec{\nabla}\phi = \nabla^2\phi = -4\pi k\rho = -\dfrac{\rho}{\epsilon_0}, \end{equation}\qquad{(3)}\]

which is an example of Poisson’s Equation and has a catalog of nifty solutions. In regions where there is no charge, \(\rho=0\), and eq. 3 becomes \[\begin{equation} \nabla^2\phi = 0, \end{equation}\qquad{(4)}\] which is an example of Laplace’s Equation and has even easier-to-use solutions, and easy-to-use numerical solution methods, where you utilize the boundary conditions to pin down your solutions, often as superposition of solutions that are orthogonal functions. Such methods occur widely in electrostatics, also in gravitation, heat and mass transfer, fluid mechanics, nuclear reactor theory, diffusion, etc.