Electric Potential

Electric potential energy is measured in Joules ($J$). Electric potential is measured in volts ($V$).

Similarly to gravitational potential energy, electric potential energy.

Review - Gravitational Potential Energy

Recall that

\[\Delta E_{mechanical} = \Delta K + \Delta U\]

where $\Delta K$ is kinetic energy and $\Delta U$ is potential energy.

Recall also that the change in potential energy is equal to to negative of the amount of work done

\[\Delta U = -W\]

and that the work done by gravity is

\[W=\vec F\cdot\vec d = mgd\cos{\theta}\]

where $m$ is mass, $g$ is the gravitational constant, $d$ is displacement, and $\theta$ is the angle between the direction of the gravitational field and the direction of displacement.

This yields the equation for gravitational potential energy

\[\Delta U_{mechanical}=mgh\]

Electric Potential Energy

In an uniform electric field, we similarly have

\[W=\vec F\cdot\vec d=qEd\cos{\theta}\]

where $q$ is the particle’s charge, $E$ the magnitude of the electric field, and $d$ the magnitude of displacement of the charged particle according to the direction of the electric field.

This yields the equation for electrical potential energy

\[U_{electric}=qEd =qV\]

And the equation for electrical potential

\[V = Ed\]

and

\[\Delta V=-Ed\]

• Voltage is electric potential
• Voltage units are volts, $1V=\frac{J}{C}=\frac{N\cdot m}{C}$
• Recall that units of electric field strength are $\frac{N}{C}$
• Moving with the field lowers electric potential, hence the minus sign $-$

Voltage of a Constant $\vec E$ field

The change in electric potential is defined as

\[\Delta V = \frac{\Delta U}{q}\]

and may be integrated from the electric field:

\[\Delta V = -\int{E_s\cdot ds}\]

where $r$ is displacement.

Hence, from the electric potential, the electric field may be differentiated:

\[E_s = -\frac{dV}{ds}\]

Voltage of a Point Charge

Recall that the magnitude of the E-field of a point charge is

\[\vec E = \frac{kq}{r^2}\]

The voltage of a point charge with charge $q$ is then

\[\Delta V = V_f-V_i= -\int_{r_i}^{r_f}{\frac{kq}{r^2}dr}\cos(\theta)\]

and assuming $\theta=0$ (the displacement is in the direction of the $E$ field) then

\[V_f-V_i = -kq\left[-\frac{1}{r}\right]_{r_i}^{r_f}=\frac{kq}{r_f}-\frac{kq}{r_i}\]

When $r_f\to\infin$, $V_f=0$ and thus

\[\bold{V_{pt.\ charge}} = 0 - V_i = -(-\frac{kq}{r}) = \bold{\frac{kq}{r}}\]

Voltage from Multiple Point Charges

We may use the principle of superposition to determine the net voltage at a point by taking the sum of the individual voltages:

\[V_{net} = V_1+V_2+\cdots+V_{n-1}+V_n\]

Converting between Voltage, Potential Electric Energy, and Kintetic Energy

1. Find voltage
2. Find potential electric energy $U_{electric}=qV$
3. Convert from potential energy to kinetic: $U=K=\frac{1}{2}mv^2$

Electric Potential of a Parallel Plate Capacitor

The electric potential energy within a parallel plate capacitor is

\[U_{electric} = qEs\]

And the electric potential is

\[V_{capacitor} = Es\]

where $s$ is the distance from the negative electrode.

In practical applications, the electric field strength $E$ of a capacitor is often defined as

\[E = \frac{\Delta V_{capacitor}}{d}\]

where $d$ is the distance between the plates.

Electric Potential Energy of a Dipole

A dipole with moment

\[\vec p = q\vec s\]

where $\vec s$ points from the negative charge to the positive charge, has electric potential energy

\[U_{electric}=-\vec p\cdot\vec E\\ = -|\vec p||\vec E|\cos\theta\]

where $\theta$ is the angle between the direction of the $\vec E$ field and the dipole moment.

Finding Potential from $E$ field

The potential difference between points $s_f$ and $s_i$ along the $s$-axis is

\[\Delta V = -\int_{s_i}^{s_f}{E_s\ ds}\]

Finding $E$ field from Potential

The component of the $E$ field in the $s$-direction is

\[E_s = -\frac{dV}{ds}\]