18.5 Electric potential

Definition of Electric Potential ($\mathrm{V}$)

Definition

The electric potential at a point is defined as the work done per unit positive charge in brining a small test charge from infinity to the point. $\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}$

Unit: Volt ($\mathrm{V}$) or ${\mathrm{J}\mathrm{C}}^{-1}$. Nature: Scalar The Reference Point: Infinity is defined as the point of zero potential.

Critical: Electric Potential is scalar

• Potential does not have to be resolved into components (vectors).
• You simply add them: ${\mathrm{V}}_{total}={\mathrm{V}}_1+{\mathrm{V}}_2+\dots$ This does not mean that potential cannot be negative!

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Electric Potential of a Point Charge

For an isolated point charge $Q$, the potential at a distance $r$ is given by:

$$\mathrm{V}=\frac{\mathrm{Q}}{4\pi {ϵ}_{\circ }r}$$

Precision: Signs Matter!

Unlike Force or Field Strength where we often calculate magnitude and decide direction later, for Potential you must substitute the sign of the charge.

Source Charge	Potential ($\mathrm{V}$)	Meaning
Positive ($+1Q$)	Positive ($-\mathrm{V}$)	Work must be done by an external agent to push a positive text charge against repulsion.
Negative ($-1Q$)	Negative ($-\mathrm{V}$)	Work is done by the field to pull a positive test charge in (Potential Well).

Graphical Representation

TODO: Add graphs

Graph

• Shape: Rectangular Hyperbola ($\mathrm{V}\propto \frac{1}{r}$).
• Comparison: It falls off slower than Field Strength ($\frac{1}{r^2}$).
• At surface ($R$): $\mathrm{V}=\frac{\mathrm{Q}}{4\pi {ϵ}_{\circ }R}$.
• Inside a Conductor ($r<R$): Potential is constant.

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Relationship between Field Strength and Potential

Relationship

Electric Field Strength is the negative potential gradient. $\mathrm{E}=-\frac{\mathrm{\Delta V}}{\Delta d}orE=-\frac{dV}{dr}$

• Negative Sign: Indicates that the field lines point in the direction of decreasing potential (High Potential $\to$ Low Potential).

Graphical Representation

TODO: Add graphs

Graph

• The Gradient of a $\mathrm{V}-\mathrm{r}$ graph is $-\mathrm{E}$.
• The Area under an $\mathrm{E}-\mathrm{r}$ graph is $\mathrm{\Delta V}$.

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Electric Potential Energy

If Electric Potential is “work done per unit charge”, then Electric Potential Energy is the total work for a specific charge $q$.

Electric Potential Energy

$\begin{array}{rl}{\mathrm{E}}_{\mathrm{p}}& =q\mathrm{\Delta V}\\ {\mathrm{E}}_{\mathrm{p}}& =\frac{Qq}{4\pi {ϵ}_{\circ }r}\end{array}$

Unit: Joule ($\mathrm{J}$). Nature: Scalar.

Determining the Nature of the Interaction

The sign of ${\mathrm{E}}_{\mathrm{p}}$ tells you the stability of the system:

1. Like Charges ($+,+$ or $-,-$): ${\mathrm{E}}_{\mathrm{p}}$ is Positive.
   • Repulsive system.
   • You must put energy in to bring the charges together.
   • The system “wants” to fly apart (converting ${\mathrm{E}}_{\mathrm{p}}\to {\mathrm{E}}_{\mathrm{k}}$).
2. Opposite Charges ($+,-$): ${\mathrm{E}}_{\mathrm{p}}$ is Negative.
   • Attractive system.
   • To separate the charges (move to $\infty$), you must supply energy (Work Done).

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Application: Conservation of Energy

Concept: Loss in ${\mathrm{E}}_{\mathrm{k}}$ = Gain in ${\mathrm{E}}_{\mathrm{p}}$

$\begin{array}{rl}\mathrm{\Delta }{\mathrm{E}}_{\mathrm{k}}& =\mathrm{\Delta }{\mathrm{E}}_{\mathrm{p}}\\ \frac{1}{2}m{v}^2& =\frac{Qq}{4\pi {ϵ}_{\circ }r}\end{array}$

Example: Distance of Closest Approach

An $\alpha$-particle (${}_2^4\text{He}$) is travelling in a vacuum directly towards the centre of a stationary gold nucleus (${}_{79}^{197}\text{Au}$), as illustrated in the diagram below

The $\alpha$-particle is emitted from a source with an initial kinetic energy of 4.8 MeV. At a large distance from the gold nucleus, the $\alpha$-particle has its maximum speed. As it approaches the nucleus, it slows down until it comes to a momentary rest at a distance $d$ from the centre of the gold nucleus.

The $\alpha$-particle and the gold nucleus may be considered to be point charges that are isolated in space.

Calculate: (a) The initial kinetic energy of the alpha particle in Joules. (b) The distance $d$.

(a) Finding the initial kinetic energy in joules $\begin{array}{rl}\mathrm{V}& =\frac{\mathrm{W}}{\mathrm{Q}}\\ \mathrm{W}& =\mathrm{Q}\mathrm{V}\\ 4.8\mathrm{MeV}& =4.8\times 10^6\times 1.6\times 10^{-19}\times 1\\ & =7.68\times 10^{-13}\mathrm{J}\end{array}$

(b) Finding the distance of closest approach, $d$ $\begin{array}{rl}\mathrm{\Delta }{\mathrm{E}}_{\mathrm{p}}& =\mathrm{\Delta }{\mathrm{E}}_{\mathrm{k}}\\ \frac{Qq}{4\pi {ϵ}_{\circ }r}& =7.68\times 10^{-13}\\ r& =\frac{79\times 2\times {\left(1.6\times 10^{-19}\right)}^2}{7.68\times 10^{-13}}\times 8.99\times 10^9\\ & =4.7\times 10^{-14}\mathrm{m}\end{array}$
$\boxed{\therefore d=4.7\times 10^{-14}\mathrm{m}}$

Tip: Mega Electron Volt ( $\mathrm{MeV}$) to Joules ($\mathrm{J}$)

$1\mathrm{MeV}=1.6\times 10^{-13}\mathrm{J}$

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Summary Comparison: Potential vs Field

Feature	Electric Field Strength ($E$)	Electric Potential ($V$)
Definition	Force per unit +ve charge	Work per unit +ve charge
Type	VECTOR (Magnitude & Direction)	SCALAR (Magnitude & Sign)
Formula (Point)	$\frac{Q}{4\pi {ϵ}_0{r}^2}$	$\frac{Q}{4\pi {ϵ}_{0}r}$
Relation	Gradient of Potential	Integral of Field
Inside Conductor	Zero	Constant (Non-zero)
Combined	Vector Addition	Algebraic Sum ($V_1+V_2$)

The "Zero" Trap

Scenario: Two equal positive charges are separated by distance $2x$.

• Midpoint Field: $\mathrm{E}=0$ (Vectors cancel: $\to$ vs $\leftarrow$).
• Midpoint Potential: $\mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2\ne 0$. (Scalars add: $\mathrm{V}+\mathrm{V}=2V$).

Scenario: One positive $+Q$ and one negative $-Q$ separated by $2x$.

• Midpoint Field: $\mathrm{E}\ne 0$ (Vectors add: $\to$ and $\to$).
• Midpoint Potential: $\mathrm{V}=0$ (Scalars cancel: $+\mathrm{V}$ and $-\mathrm{V}$).