Potential Difference and Potential Energy

1. Introduction

When a charge is placed in an electric field it experiences a force, so work has to be done to move it from one point to another. This work is stored as electric potential energy, and the energy per unit charge defines two closely related quantities: electric potential and potential difference. These ideas let us describe an electric field in terms of energy (a scalar) rather than force (a vector), which is often far simpler.

Throughout this chapter the medium is assumed to be free space (vacuum or air), so the constant that appears is $\dfrac{1}{4\pi\varepsilon_0}$, where $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space. A useful numerical value is:

$$\frac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N m}^2\text{C}^{-2}$$

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2. Electric Potential Energy

The electric potential energy of a charge at a point in an electric field is the work done in bringing that charge from infinity (where the field has no effect) to the point, against the electrostatic force.

Because the electrostatic force is conservative, this work depends only on the end position, not on the path taken. If $W$ is the work done in bringing a charge $q_0$ from infinity to a point, the potential energy at that point is $U = W$, measured in joules ($\text{J}$).

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3. Electric Potential

3.1 Definition (Subtopic 21.1)

The electric potential at a point in an electric field is defined as the work done per unit positive charge in bringing a small positive test charge from infinity to that point against the electrostatic force.

If $W$ is the work done in bringing a small positive test charge $q_0$ from infinity to a point, the potential at that point is:

$$V = \frac{W}{q_0}$$

Potential is a scalar quantity. Its SI unit is the volt ($\text{V}$), where $1\ \text{V} = 1\ \text{J C}^{-1}$.

The test charge is taken to be small so that it does not disturb the field whose potential is being measured.

3.2 Potential due to a point charge (Subtopic 21.3)

Consider a point charge $+q$ placed at a point $O$. We find the potential at a point $A$ at distance $r$ from the charge by carrying a unit positive test charge from infinity to $A$.

Let the test charge be at an intermediate point $P$ at distance $x$ from $q$. The force on the unit positive charge at $P$ is, by Coulomb's law:

$$F = \frac{1}{4\pi\varepsilon_0},\frac{q}{x^2}$$

To move the charge a small displacement $dx$ towards $q$ (that is, in the direction of decreasing $x$), the work done against the electrostatic force is:

$$dW = -F,dx = -\frac{1}{4\pi\varepsilon_0},\frac{q}{x^2},dx$$

The negative sign appears because the displacement $dx$ (measured outward, in the $+x$ direction) is opposite to the direction in which the charge is actually moved. The total work done in bringing the unit charge from infinity ($x = \infty$) to $A$ ($x = r$) is:

$$ \begin{aligned} W &= -\frac{q}{4\pi\varepsilon_0}\int_{\infty}^{r}\frac{dx}{x^2} \ &= -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{x}\right]_{\infty}^{r} \ &= -\frac{q}{4\pi\varepsilon_0}\left(-\frac{1}{r}+0\right) \ W &= \frac{q}{4\pi\varepsilon_0},\frac{1}{r} \end{aligned} $$

Since the potential is the work done per unit positive charge, the potential at $A$ is:

$$\boxed{,V = \frac{1}{4\pi\varepsilon_0},\frac{q}{r} = \frac{q}{4\pi\varepsilon_0 r},}$$

Features of this result:

1. For a positive charge $V$ is positive, and for a negative charge $V$ is negative.
2. $V \propto \dfrac{1}{r}$: the potential falls off with distance, becoming zero at infinity.
3. Potential is a scalar, so the potential due to several point charges at a point is the algebraic sum of the individual potentials: $V = \dfrac{1}{4\pi\varepsilon_0}\sum \dfrac{q_i}{r_i}$.

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