Chapter 4: The Electric Potential – Study Notes

Electric Potential

Electrical Potential Energy

When a charge q moves in a constant electric field E, it experiences a force F = qE. The work done by the field as the charge moves from point r1 to r2 is given by:

• The work done: $W = \int_{r_1}^{r_2} \mathbf{F} \cdot d\mathbf{s}$

• For the electrostatic force: $W = \int_{r_1}^{r_2} q\mathbf{E} \cdot d\mathbf{s} = q \int_{r_1}^{r_2} \mathbf{E} \cdot d\mathbf{s}$

• The electric force is conservative, so the work done does not depend on the path taken.

• The change in potential energy is the negative of the work done by the electric force: $\Delta U = -W = -q \int_{r_1}^{r_2} \mathbf{E} \cdot d\mathbf{s}$

• Potential energy is usually defined to be zero at infinity: $U_{\infty} = 0$.

Electric Potential

The electric potential V is defined as the potential energy per unit charge. It is a scalar quantity and is useful for calculating energy changes for any charge.

• Definition: $\Delta V = \frac{\Delta U}{q}$

• Potential difference between two points: $\Delta V = -\int_{r_1}^{r_2} \mathbf{E} \cdot d\mathbf{s}$

• Units: $1\ \mathrm{V} = 1\ \mathrm{J}/\mathrm{C}$ (volt = joule per coulomb)

• Energy from charge and potential difference: $U = q\Delta V$

• Electron-volt: $1\ \mathrm{eV} = (e)(1\ \mathrm{V}) = 1.60 \times 10^{-19}\ \mathrm{J}$

• To define $V(r)$ at all points, specify a reference point (often $V = 0$ at infinity): $V(r) = -\int_{r_{\mathrm{ref}}}^{r} \mathbf{E} \cdot d\mathbf{s}$

Equipotential Surfaces

An equipotential surface is a set of points where the electric potential has the same value. No work is required to move a charge along an equipotential surface ($\Delta V = 0$).

• Electric field lines are always perpendicular to equipotential surfaces.

Finding E from V

The electric field can be found from the electric potential by taking derivatives:

• $E_x = -\frac{\partial V}{\partial x}$

• $E_y = -\frac{\partial V}{\partial y}$

• $E_z = -\frac{\partial V}{\partial z}$

• Or, in vector notation: $\mathbf{E} = -\nabla V$

Potential of a Point Charge and Groups of Point Charges

The electric potential due to a point charge q at a distance r (with $V = 0$ at infinity):

• $V(r) = k\frac{q}{r} = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$

• For multiple point charges: $V(r) = \sum_i k\frac{q_i}{r_i} = \sum_i \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$

• For an electric dipole at the origin (along the z-axis): $V(r) = \frac{1}{4\pi\varepsilon_0} \frac{p \cos \theta}{r^2}$

Potential Due to a Continuous Charge Distribution

For a continuous charge distribution with charge density $\rho(r)$:

• Potential due to a small element $dq$ at distance $r$: $dV = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r}$

• Total potential: $V = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{r} = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(r) d\tau}{r}$

Potential Energy of a System of Charges

The potential energy of two point charges $q_1$ and $q_2$ separated by $r$:

• $U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$

• For a system of charges: $U = \frac{1}{4\pi\varepsilon_0} \sum_{\text{pairs } ij} \frac{q_i q_j}{r_{ij}}$ Each pair is counted only once.

Worked Examples

Electric Potential Calculations

1. Energy Change of an Electron in a Thunderstorm Given: $\Delta V = 1.2 \times 10^9\ \mathrm{V}$ Find: Change in energy for an electron.

   • $|\Delta U| = |q\Delta V| = e(1.2 \times 10^9\ \mathrm{V}) = 1.2 \times 10^9\ \mathrm{eV} = 1.2\ \mathrm{GeV}$

2. Equipotential Surfaces for a Charged Sheet Given: $\sigma = 0.10\ \mu\mathrm{C}/\mathrm{m}^2$, $\Delta V = 50\ \mathrm{V}$ Find: Distance between equipotential surfaces.

   • Electric field: $E_z = \frac{\sigma}{2\varepsilon_0} = 5.64 \times 10^3\ \mathrm{N}/\mathrm{C}$

   • $\Delta V = -E_z \Delta z \implies \Delta z = -\frac{\Delta V}{E_z} = -8.8 \times 10^{-3}\ \mathrm{m} = -8.8\ \mathrm{mm}$

3. Potential Difference Between Parallel Plates Given: $d = 0.12\ \mathrm{m}$, $F = 3.9 \times 10^{-15}\ \mathrm{N}$ on an electron Find: (a) $E$, (b) $\Delta V$

   • $E = \frac{F}{e} = 2.4 \times 10^4\ \mathrm{N}/\mathrm{C}$

   • $|\Delta V| = E d = (2.4 \times 10^4\ \mathrm{V}/\mathrm{m})(0.12\ \mathrm{m}) = 2.9 \times 10^3\ \mathrm{V}$

4. Potential Inside a Nonconducting Sphere Given: Uniform charge $q$ in sphere of radius $R$, $E(r) = \frac{q r}{4\pi\varepsilon_0 R^3}$ Find: $V(r)$ inside, with $V(0) = 0$.

   • $V(r) = -\int_0^r E(r') dr' = -\frac{q}{4\pi\varepsilon_0 R^3} \int_0^r r' dr' = -\frac{q r^2}{8\pi\varepsilon_0 R^3}$

   • Potential difference between surface and center: $V(R) - V(0) = -\frac{q}{8\pi\varepsilon_0 R}$

   • If $q > 0$, the center is at higher potential.

5. Potential Inside a Uniformly Charged Sphere (Reference at Infinity) Given: $V = 0$ at $r = \infty$, $r < R$ Find: $V(r)$

   • Split integral for inside and outside:

   • $V(r) = \frac{q(3R^2 - r^2)}{8\pi\varepsilon_0 R^3}$

   • Potential difference between surface and center is the same as previous example.

6. Charge and Charge Density on a Conducting Sphere Given: $R = 0.15\ \mathrm{m}$, $V = 200\ \mathrm{V}$ at surface, $V = 0$ at infinity Find: (a) $Q$, (b) $\sigma$

   • $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R} \implies Q = 4\pi\varepsilon_0 V R = 3.3 \times 10^{-9}\ \mathrm{C}$

   • $\sigma = \frac{Q}{4\pi R^2} = 1.2 \times 10^{-8}\ \mathrm{C}/\mathrm{m}^2$

7. Potential at Center of a Hollow Conducting Sphere Given: $V = 400\ \mathrm{V}$ at surface, $Q = 5.0 \times 10^9\ \mathrm{C}$ Find: $V$ at center

   • Inside a conductor, $E = 0$, so $V$ is constant throughout. $V_{\text{center}} = 400\ \mathrm{V}$

8. Excess Charge on a Conducting Sphere Given: $R = 0.15\ \mathrm{m}$, $V = 1500\ \mathrm{V}$ at surface Find: $Q$

   • $Q = R V / k = (0.15\ \mathrm{m})(1500\ \mathrm{V}) / (8.99 \times 10^9\ \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2) = 2.5 \times 10^{-8}\ \mathrm{C}$

9. Electric Field from a Potential Function Given: $V = (2.0\ \mathrm{V}/\mathrm{m}^2)x^2 - (3.0\ \mathrm{V}/\mathrm{m}^2)y^2$, at $(3.0, 2.0)$ m Find: $\mathbf{E}$

   • $E_x = -\frac{\partial V}{\partial x} = -4.0x = -12\ \mathrm{V}/\mathrm{m}$

   • $E_y = -\frac{\partial V}{\partial y} = +6.0y = 12\ \mathrm{V}/\mathrm{m}$

   • $|\mathbf{E}| = \sqrt{(-12)^2 + (12)^2} = 17\ \mathrm{V}/\mathrm{m}$

   • Direction: $\theta = \tan^{-1}(1) = 135^\circ$ (second quadrant)

Potential Energy of a System of Charges

10. Potential Energy of Two Electrons Given: $r = 2.00\ \mathrm{nm}$ Find: $U$

    • $U = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} = 1.15 \times 10^{-19}\ \mathrm{J}$

    • As $r$ increases, $U$ decreases (since $U \propto 1/r$)

11. Work to Assemble Four Charges in a Square Given: Four charges $+q$, $-q$, $-q$, $+q$ at corners of a square of side $a$ Find: Total work to assemble configuration

    • Sum work for each step (see detailed derivation in content)

    • Total work: $W_{\text{Total}} = \frac{q^2}{4\pi\varepsilon_0 a} \left(-4 + 2\sqrt{2}\right)$

12. Potential and Work in a Rectangle of Charges Given: Rectangle sides $5.0$ cm and $15$cm,$q_1 = -5.0\ \mu\mathrm{C}$,$q_2 = +2.0\ \mu\mathrm{C}$ Find: (a) $V_A$, (b) $V_B$, (c) Work to move $q_3 = +3.0\ \mu\mathrm{C}$ from $B$ to $A$, (d) Path dependence

    • (a) $V_A = 6.0 \times 10^4\ \mathrm{V}$

    • (b) $V_B = -7.8 \times 10^5\ \mathrm{V}$

    • (c) $\Delta V = 8.4 \times 10^5\ \mathrm{V}$, $\Delta U = 2.5\ \mathrm{J}$

    • (d) Work is independent of path (conservative force)

13. Charged Spheres Connected by a String Given: $m_A = 5.00\ \mathrm{g}$, $m_B = 10.0\ \mathrm{g}$, $q = 5.00\ \mu\mathrm{C}$, $d = 1.00\ \mathrm{m}$ Find: (a) $U$, (b) Accelerations after string is cut, (c) Final speeds

    • (a) $U = 0.225\ \mathrm{J}$

    • (b) $F = 0.225\ \mathrm{N}$, $a_A = 45.0\ \mathrm{m}/\mathrm{s}^2$, $a_B = 22.2\ \mathrm{m}/\mathrm{s}^2$

    • (c) Conservation of energy and momentum yields $v_A = 7.75\ \mathrm{m}/\mathrm{s}$, $v_B = 3.87\ \mathrm{m}/\mathrm{s}$

14. Electron Shot Between Two Fixed Electrons Given: Two electrons fixed $2.0$ cm apart, third electron shot from infinity, stops midway Find: Initial speed $v_0$

    • Potential energy increase: $\Delta U = \frac{e^2}{2\pi\varepsilon_0 r}$, $r = 1.0\ \mathrm{cm}$

    • Energy conservation: $\frac{1}{2} m v_0^2 = \Delta U$

    • $v_0 = 3.18 \times 10^2\ \mathrm{m}/\mathrm{s}$

Summary Table: Key Equations

Concept	Equation	Description
Work by Electric Field	$W = \int \mathbf{F} \cdot d\mathbf{s}$	Work done moving charge in field
Potential Difference	$\Delta V = -\int \mathbf{E} \cdot d\mathbf{s}$	Potential difference between two points
Electric Field from Potential	$\mathbf{E} = -\nabla V$	Field as gradient of potential
Potential of Point Charge	$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$	Potential at distance $r$ from charge $q$
Potential Energy (2 charges)	$U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$	Energy of two point charges
Potential (Continuous Distribution)	$V = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(r) d\tau}{r}$	Potential from charge density
Electron-volt	$1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}$	Energy unit

Additional info: Some explanations and step-by-step derivations have been expanded for clarity and completeness. All equations are provided in LaTeX format for clarity and further study.