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Section 4.1 Electrostatic and Gravitational Potentials and Potential Energies

Recall that the electrostatic potential \(V\) due to a point charge \(q\) stationary at the origin is given by

\begin{equation} V = \frac{1}{4\pi\epsilon_0} \, \frac{q}{r}\tag{4.1.1} \end{equation}

where \(r\) is the distance from the point charge to the place at which the potential is being evaluated. Similarly, the gravitational potential \(\Phi\) due to a point mass \(m\) is given by

\begin{equation} \Phi = -G \, \frac{m}{r} .\tag{4.1.2} \end{equation}

In the coming sections of the text, we will be discussing both the physical meaning of these potentials and how to calculate them in a number of different situations:

  • For multiple (discrete) charges;

  • For idealized continuous charge distributions;

  • For symmetric charge distributions;

  • Using approximation methods;

  • In limiting situations.

The constant \(\epsilon_0\) is called the permittivity of free space. The constant \(G\) is called Newton's constant. In SI units, the values of these constants are:

\begin{align*} \epsilon_0 \amp = 8.85\times10^{-12}\frac{ C^2}{ Nm^2}\\ G \amp = 6.67\times10^{-11}\frac{ Nm^2}{ kg^2} \end{align*}

The potential energy experienced by a test charge \(q_{ test}\) in an electrostatic potential is given by \(q_{ test}V\text{.}\) Therefore the potential is the potential energy per unit charge.

\begin{align*} V \amp \sim \frac{ ML^2}{T^2}\frac{1}{C}\amp\text{Energy per unit charge} \end{align*}

The test charge, if unconstrained, will move so as to minimize the potential energy.

Question 4.1.1. Electrostatic vs. Gravitational Potentials.

(a)

Find the dimensions of gravitational potential and compare to electrostatic potential.

Answer.

The potential energy experienced by a test mass \(m_{ test}\) in a gravitational potential is given by \(m_{ test}\Phi\text{.}\) Therefore the potential is the potential energy per unit mass.

\begin{align*} \Phi \amp \sim \frac{ ML^2}{ T^2}\frac{1}{M}\amp\text{Energy per unit mass} \end{align*}

compared to

\begin{align*} V \amp \sim \frac{ ML^2}{T^2}\frac{1}{C}\amp\text{Energy per unit charge} \end{align*}

i.e.  the dimensions of potential energy are the same in both physical situations, but the dimensions of potential are different.

(b)

Use the formulas for electrostatic and gravitational potential energies to explain why two masses attract each other but two like charges repel.

Hint.

The negative sign in the gravitational potential is important.

(c)

Find the dimensions (see Section 2.1) for \(\epsilon_0\) and \(G\text{?}\)

Answer.

It may be easiest to use the formula for kinetic energy \(\frac{1}{2}mv^2\) to quickly infer the dimensions of energy \(E\sim \frac{ML^2}{T^2}\text{.}\) Remember that potentials are energy per unit charge (\(V\)) or energy per unit mass (\(\Phi\)). Then you know the dimensions of every symbol in Equation (4.1.1) and Equation (4.1.2) except \(\epsilon_0\) and \(G\text{,}\) so you can solve for these dimensions using straightforward algebra:

\begin{align*} \epsilon_0 \amp \sim \frac{ C^2}{ M\frac{L}{T^2}L^2} = \frac{ C^2T^2}{ ML^3} ,\\ G \amp \sim \frac{ M\frac{L}{T^2}L^2}{ M^2} = \frac{ L^3 }{ MT^2} . \end{align*}