Electric Potential and Potential Energy

Electrostatic potential energy

• The potential energy of a system is the stored energy that depends on the relative positions of its constituents. Electrostatic potential energy is the work done against the electrostatic forces to achieve a certain configuration of charges.
• Since every system tries to attain the lowest potential energy, work has to be done to change the configuration. Work done against the electrostatic force equals the increase in the potential energy of the system: $\vec{F}\cdot d\vec{r} = dU$.

Expression for the potential energy of two point charges

Set-up

• Let a source charge $+Q$ be fixed at the origin O. A small charge $+q_0$ is brought from point A to point B, at distances $r_1$ and $r_2$ from O, against the repulsive force on it.
• The Coulomb force on $q_0$ at distance $r$ is

$$\vec{F}_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq_0}{r^2}\hat{r}$$

Derivation

• The work done against $\vec{F}_E$ in a small displacement $d\vec{r}$ appears as an increase in potential energy, $dU = -\vec{F}_E\cdot d\vec{r}$, the negative sign appearing because $d\vec{r}$ is directed against the force.
• Integrating from A to B,

$$\Delta U = -\int_{A}^{B}\vec{F}_E\cdot d\vec{r} = -\dfrac{Qq_0}{4\pi\varepsilon_0}\int_{r_1}^{r_2}\dfrac{dr}{r^2}$$

$$\Delta U = \dfrac{Qq_0}{4\pi\varepsilon_0}\left[\dfrac{1}{r}\right]_{r_1}^{r_2} = \dfrac{Qq_0}{4\pi\varepsilon_0}\left(\dfrac{1}{r_2}-\dfrac{1}{r_1}\right)$$

• The change depends only on the end points, not on the path taken; it equals the work done $W_{AB}$ against the electrostatic force.
• Choosing infinity as the point of zero potential energy (the force is zero at $r=\infty$), put $r_1 = \infty$ and $r_2 = r$ for two charges $q_1$ and $q_2$:

$$\boxed{\,U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}\,}$$

Units

• SI unit of potential energy is the joule: one joule is the energy in moving a charge of 1 C through a potential difference of 1 volt.
• The electron volt (eV) is the change in kinetic energy of an electron crossing two points at a potential difference of 1 volt: $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. Also $1\ \text{keV} = 1.6\times10^{-16}\ \text{J}$ and $1\ \text{meV} = 1.6\times10^{-22}\ \text{J}$.

Concept of potential

• Write the two-charge energy as $U = q_2\left(\dfrac{q_1}{4\pi\varepsilon_0 r}\right) = q_1\left(\dfrac{q_2}{4\pi\varepsilon_0 r}\right)$. The bracketed quantity $V(r) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ depends only on the charge $q$ and the distance $r$, and is defined as the electrostatic potential of the charge $q$ at that distance.
• Hence electrostatic potential energy = electric potential $\times$ charge, or electrostatic potential is potential energy per unit charge:

$$V = \dfrac{U}{q}$$

• The potential difference between two points is

$$V_2 - V_1 = \dfrac{U_2-U_1}{q_0} = \dfrac{W}{q_0}$$

• i.e. the work done per unit charge in moving the charge between the two points. Potential is a scalar, measured in volt.

Relation between electric field and electric potential

• Consider the field of a charge $+q$ at O. Move a unit positive charge from M to N, a small distance $dx$ from M along the field direction. The work done is $dW = -F\,dx = -E\,dx$, the negative sign because the charge is moved against the force.
• As this is work on unit positive charge, $dW = dV$, so

$$dV = -E\,dx\qquad\Rightarrow\qquad \boxed{\,E = -\dfrac{dV}{dx}\,}$$

• The electric field at a point is the negative gradient of the potential at that point. For plates a distance $d$ apart at a potential difference $V$, the magnitude is $E = \dfrac{V}{d}$.

Zero of potential

• The zero point of potential is arbitrary, but once chosen, every potential is measured with respect to it.
• For a point charge or a localised collection of charges, the zero is set at infinity. For electrical circuits, the earth is taken as zero.
• With that convention, the potential at a point A is the work done to bring a unit positive charge from infinity to A.

Electric potential due to a point charge

Derivation

• Let $+q$ be at O and A be at distance $r$. Since the work is path-independent, choose the path along OA extended to infinity. At an intermediate point M with $OM = x$, the force on a unit positive charge is

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

• directed away from O. For an infinitesimal displacement $dx$ from M towards O, $dW = -F\,dx$. Total work in bringing unit positive charge from infinity to A:

$$W = -\int_{\infty}^{r}F\,dx = -\dfrac{q}{4\pi\varepsilon_0}\int_{\infty}^{r}\dfrac{dx}{x^2} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{x}\right]^{r}_{\infty}$$

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

Reading the result

• A positive charge produces a positive potential and a negative charge a negative potential. At $r=\infty$, $V=0$, so potential is zero at infinity.
• $V$ has the same value at all points at distance $r$, whatever the direction, so the potential of a single charge is spherically symmetric.
• Compare the fall-offs: $F\propto\dfrac{1}{r^2}$, $E\propto\dfrac{1}{r^2}$, but $V\propto\dfrac{1}{r}$.

Electric potential due to an electric dipole

Set-up and diagram

• A dipole AB consists of $+q$ at A and $-q$ at B separated by $2l$, with dipole moment $p = q\times 2l$ directed from $-q$ to $+q$. The line joining the charges is the axis; the perpendicular through the centre O is the equator.
• Let C be a point at distance $r$ from O, inclined at angle $\theta$ to the axis, with $r_1 = AC$ and $r_2 = BC$. The diagram shows the two charges, the axis, the angle $\theta$ at O, and the two distances $r_1$ and $r_2$ from the charges to C.

Derivation

• Potentials at C due to the two charges are $V_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r_1}$ and $V_2 = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r_2}$, so

$$V_C = V_1+V_2 = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$$

• By geometry, $r_1^2 = r^2 + l^2 - 2rl\cos\theta$ and $r_2^2 = r^2 + l^2 + 2rl\cos\theta$. For a short dipole $2l \ll r$, so $\dfrac{l^2}{r^2}$ is neglected:

$$r_1 = r\left(1 - \dfrac{2l\cos\theta}{r}\right)^{1/2},\qquad r_2 = r\left(1 + \dfrac{2l\cos\theta}{r}\right)^{1/2}$$

• Using the binomial expansion $(1+x)^n \approx 1+nx$ for $x\ll 1$ and keeping only first-order terms in $\dfrac{l}{r}$,

$$\dfrac{1}{r_1} = \dfrac{1}{r}\left(1+\dfrac{l\cos\theta}{r}\right),\qquad \dfrac{1}{r_2} = \dfrac{1}{r}\left(1-\dfrac{l\cos\theta}{r}\right)$$

$$V_C = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{2l\cos\theta}{r^2}$$

• Since $p = q\times 2l$,

$$\boxed{\,V_C = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}\,}$$

Special positions

• Axial point, $\theta = 0^\circ$ (towards $+q$) or $180^\circ$ (towards $-q$): $V_{\text{axial}} = \pm\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$, the maximum magnitude.
• Equatorial point, $\theta = 90^\circ$: $\cos\theta = 0$, so $V = 0$. The plane perpendicular to the dipole axis through its midpoint is an equipotential plane of zero potential, and no work is done in moving a charge anywhere in it.

Electrostatic potential due to a system of charges

• For charges $q_1, q_2,\ldots,q_n$ at distances $r_1, r_2,\ldots,r_n$ from a point P, each contributes $V_i = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_i}$.
• By the principle of superposition the total potential at P is the algebraic sum of the individual potentials:

$$\boxed{\,V = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}+\cdots+\dfrac{q_n}{r_n}\right) = \dfrac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\dfrac{q_i}{r_i}\,}$$

• For a continuous charge distribution the summation is replaced by integration. Because potential is a scalar, the signs of the charges do all the work — never resolve components.