FAD1022 L5 — Electric Flux and Gauss Law (continued)

Course: FAD 1022 – Basic Physics 2
Semester: 2 2025/2026
Lecturer: Dr. Hashlina Rusdi
Topics: Gauss's Law Applications, Conductors, Electric Potential

Summary

This lecture continues with applications of Gauss's Law, focusing on conductors in electrostatic equilibrium, spherical charge distributions, and introduces electric potential and potential energy due to point charges.

01 — Review of Gauss's Law Applications

Field of a Line Charge (Review)

$$E = \frac{\lambda}{2\pi r \varepsilon_0}$$

  • $\vec{E}$ parallel to curved surface of Gaussian cylinder
  • No flux through end caps ($\vec{E}$ perpendicular to $\vec{A}$)

Infinite Plane Sheet (Review)

$$E = \frac{\sigma}{2\varepsilon_0}$$

Parallel Plates (Review)

$$E_{between} = \frac{\sigma}{\varepsilon_0}, \quad E_{outside} = 0$$

02 — Conductors and Electric Fields

Key Property

$E = 0$ at all points inside a conductor in electrostatic equilibrium.

Charge on a Conductor

For a Gaussian surface just inside the conductor surface:

  • Since $E = 0$ inside, flux through Gaussian surface = 0
  • Therefore, enclosed charge = 0
  • All excess charge resides on the outer surface

Field Just Outside a Conductor

Using Gauss's law with a pillbox Gaussian surface: $$E = \frac{\sigma}{\varepsilon_0}$$

The field is perpendicular to the surface.

Summary for Charged Conductor

Location Electric Field
Inside conductor $E = 0$
Just outside surface $E = \frac{\sigma}{\varepsilon_0}$
Far from conductor Like point charge

03 — Worked Examples (Gauss's Law)

Example 5: Finding Enclosed Charge

Given: $\Phi_E = 1.4$ N·m²/C
Find: $Q_{enclosed}$

$$Q = \varepsilon_0 \Phi_E = (8.85 \times 10^{-12})(1.4) = 1.24 \times 10^{-11} \text{ C}$$

Example 6: Finding Flux from Charge

Given: $Q = 2.5 \times 10^{-6}$ C
Find: Net electric flux

$$\Phi_E = \frac{Q}{\varepsilon_0} = \frac{2.5 \times 10^{-6}}{8.85 \times 10^{-12}} = 2.82 \times 10^{5} \text{ N·m}^2\text{/C}$$

Example 7: Flux Through Cube Face

Given: Point charge $q = 8.00 \times 10^{-9}$ C at center of cube with side 0.200 m
Find: Flux through one face

Total flux through cube: $$\Phi_{total} = \frac{q}{\varepsilon_0}$$

Flux through one face (6 faces): $$\Phi_{face} = \frac{q}{6\varepsilon_0} = \frac{8.00 \times 10^{-9}}{6(8.85 \times 10^{-12})} = 151 \text{ N·m}^2\text{/C}$$

Example 8: Surface Charge Density from Field

Given: $E = 4.5$ N/C from infinite plane sheet
Find: $\sigma$

$$E = \frac{\sigma}{2\varepsilon_0} \Rightarrow \sigma = 2\varepsilon_0 E = 2(8.85 \times 10^{-12})(4.5) = 7.97 \times 10^{-11} \text{ C/m}^2$$

Example 9: Charge on Parallel Plates

Given: $E = 7.20 \times 10^3$ N/C between plates, $A = 100$ cm² = $10^{-2}$ m²
Find: Charge on each plate

$$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$

$$Q = \varepsilon_0 A E = (8.85 \times 10^{-12})(10^{-2})(7.20 \times 10^3) = 6.37 \times 10^{-10} \text{ C}$$

Example 10: Distance from Line Charge

Given: $\lambda = 3.00 \times 10^{-12}$ C/m, $E = 0.600$ N/C
Find: Distance $r$

$$E = \frac{\lambda}{2\pi r \varepsilon_0} \Rightarrow r = \frac{\lambda}{2\pi \varepsilon_0 E}$$

$$r = \frac{3.00 \times 10^{-12}}{2\pi (8.85 \times 10^{-12})(0.600)} = 0.090 \text{ m} = 9.0 \text{ cm}$$

Example 11: Spherical Conducting Shells

Setup: Sphere (radius $a$, charge $q$) inside hollow sphere (inner radius $b$, outer radius $c$, no net charge)

Region Electric Field Explanation
$r < a$ $E = 0$ Inside conductor
$a < r < b$ $E = \frac{kq}{r^2}$ Outside point charge
$b < r < c$ $E = 0$ Inside conductor
$r > c$ $E = \frac{kq}{r^2}$ Net charge = $q$

Charge on inner surface of hollow sphere: $-q$ (induced)
Charge on outer surface of hollow sphere: $+q$

Example 12: Point Charge Inside Charged Shell

Setup: Point charge $+q$ at center of spherical shell (radius $a$, charge $-q$ on surface)

(a) Outside the shell ($r > a$): $$E = \frac{k(q - q)}{r^2} = 0$$

(b) Inside the shell ($r < a$): $$E = \frac{kq}{r^2}$$

Example 13: Spherical Shell with Measured Field

Given: Spherical shell with $r = 0.750$ m, $E = 890$ N/C radially inward

(a) Net charge: $$E = \frac{k|Q|}{r^2} \Rightarrow Q = -\frac{Er^2}{k} = -\frac{(890)(0.750)^2}{8.99 \times 10^9} = -5.56 \times 10^{-8} \text{ C}$$

(b) Nature of charge: Negative charge distributed uniformly on the surface (or throughout if solid conductor).

Example 14: Line Charge Field (Review)

$$E = \frac{\lambda}{2\pi r \varepsilon_0}$$

04 — Electric Potential Due to Point Charges

Definition

Electric potential ($V$) is the potential energy per unit charge.

Potential from Point Charge

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

Where:

  • $k = 8.99 \times 10^9$ N·m²/C²
  • $Q$ = source charge
  • $r$ = distance from charge

Unit: Volt (V)

1 V = 1 J/C

Properties

  • Scalar quantity — much easier to work with than vector fields
  • Positive charges create positive potential
  • Negative charges create negative potential

05 — Electric Potential Energy

Definition

Work needed to bring a charge from infinity to a point in the field.

For Point Charges

$$U = qV = \frac{kqQ}{r}$$

06 — Key Advantages of Using Potential

  1. Scalar vs Vector: Potential is a scalar — no vector components to calculate
  2. Superposition: Simply add potentials algebraically
  3. Easier calculations: For multiple charges, sum of scalars vs vector addition

07 — Worked Examples (Potential)

Example 1: Potential from Proton

Given: Proton ($Q = +e = 1.6 \times 10^{-19}$ C)

(a) At $r = 1.00$ cm: $$V = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(1.6 \times 10^{-19})}{0.01} = 1.44 \times 10^{-7} \text{ V}$$

(b) Potential difference between $r = 1.00$ cm and $r = 2.00$ cm: $$\Delta V = V_1 - V_2 = kQ\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = (8.99 \times 10^9)(1.6 \times 10^{-19})\left(\frac{1}{0.01} - \frac{1}{0.02}\right)$$ $$\Delta V = 7.19 \times 10^{-8} \text{ V}$$

Example 2: Potential from Multiple Charges

Find potential at point P due to rectangular arrangement of charges.

Approach: $$V_P = \sum_i \frac{kq_i}{r_i}$$

Sum the scalar potentials from each charge.

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