Electronic Potential

Electric Potential

Just like a mass has potential energy in a gravitational field, an electric charge has electrostatic potential energy in an electrostatic field. This energy is associated with interacting charges.

The electric potential is useful as an alternative to the electric field in solving electrostatic problems. This section covers the concept of electric potential, the potential due to charge distribution, and the motion of a charged particle in a uniform electric field.

The concept of electric potential

When a positive test charge $q$ is moved against an electric field, work is required to overcome electrostatic repulsion. The work done in moving a unit positive test charge is called electric potential:

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

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

$V = \frac{kQ}{r}$

The work done in moving a test charge $q_0$ from point B to A:

$W_{BA} = - \int_{r_B}^{r_A} F \cdot dr = -kQq_0 \int_{r_B}^{r_A} \frac{1}{r^2} dr = kQq_0 \left( \frac{1}{r_A} - \frac{1}{r_B} \right)$

So, the potential difference is:

V_A - V_B = \frac{W_{BA}}{q_0} = kQ \left( \frac{1}{r_A} - \frac{1}{r_B} \right) \tag{9.14}

If $r_B \to \infty$, the potential at point A becomes:

V_A = \frac{kQ}{r_A} \tag{9.15}

The unit of electric potential is volt (V) or $\text{JC}^{-1}$.

Electrostatic potential energy:

U = qV \quad \Rightarrow \quad U_A = qV_A = qW_{\infty A} \tag{9.16}

\therefore U = q\Delta V \tag{9.17}

Example 9

Two point charges $+20\ \mu\text{C}$ and $-20\ \mu\text{C}$ are 20 cm apart. Find the potential at a point midway.

$V_p = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} = \frac{k}{0.1} (20 \times 10^{-6} + (-20 \times 10^{-6})) = 0\ \text{V}$

Example 10

Three charges $2\ \mu\text{C}, -3\ \mu\text{C}, 4\ \mu\text{C}$ are placed at corners of an equilateral triangle of side 2 m. Find potential at the midpoint of AB.

$V_p = k\left(\frac{Q_A}{r_A} + \frac{Q_B}{r_B} + \frac{Q_C}{r_C}\right) = 9 \times 10^9 (\frac{2 \times 10^{-6}}{1} + \frac{-3 \times 10^{-6}}{1} + \frac{4 \times 10^{-6}}{3}) = 11784.6\ \text{V}$

Example 11

Two positive charges $10\ \mu\text{C}$ and $8\ \mu\text{C}$ are 10 cm apart. Find the work done in reducing the separation to 6 cm.

$\Delta V = kQ \left(\frac{1}{r_A} - \frac{1}{r_B}\right) = 9 \times 10^9 \cdot 8 \times 10^{-6} (\frac{1}{0.06} - \frac{1}{0.10}) = 4.8 \times 10^6\ \text{V}$

$\Delta W = q\Delta V = 10 \times 10^{-6} \cdot 4.8 \times 10^6 = 4.8\ \text{J}$

Example 12

Two charges $3\ \text{nC}$ and $-3\ \text{nC}$ are 3 cm apart. A dust particle with $5\ \mu\text{g}$ and $2\ \mu\text{C}$ starts from rest. Find its speed between the points.

$V_a = 1350\ \text{V},\ V_b = -1350\ \text{V},\ V_a - V_b = 2700\ \text{V}$

$\frac{1}{2}mv^2 = q(V_a - V_b) \Rightarrow v = \sqrt{\frac{2q(V_a - V_b)}{m}} = \sqrt{\frac{2 \cdot 2 \times 10^{-6} \cdot 2700}{5 \times 10^{-6}}} \approx 1469.69\ \text{ms}^{-1}$

Relationship between electric field and potential

dW_{ab} = q_0 \vec{E} \cdot d\vec{r} \Rightarrow \Delta V = V_b - V_a = - \int_a^b \vec{E} \cdot d\vec{r} \tag{9.18}

$dV = -\vec{E} \cdot d\vec{r} \Rightarrow E = -\frac{dV}{dr}$

Electric potential due to charge distribution

a. Infinite line of charge

For a linear charge density $\lambda$, electric field:

$E = \frac{\lambda}{2\pi\epsilon_0 r}$

Then potential difference:

V_a - V_b = -\int_{r_b}^{r_a} \frac{\lambda}{2\pi\epsilon_0 r} dr = -\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_b}{r_a}\right) \tag{9.19}

b. Finite line of charge

For total charge $Q$ along length $2a$, linear density $\lambda = \frac{Q}{2a}$:

$V_p = k\lambda \int_{-a}^a \frac{dy}{\sqrt{y^2 + x^2}} = \frac{\lambda}{4\pi\epsilon_0} \ln \left( \frac{a + \sqrt{a^2 + x^2}}{-a + \sqrt{a^2 + x^2}} \right)$