FISIKA Kelas 12 - Hukum Gauss & Potensial Listrik | GIA Academy

Main ideas & concepts conveyed

• Everyday connection to Gauss’s Law

  • The video starts with a balloon-on-hair example to introduce how electric effects appear in daily life.
  • It states that this phenomenon is an application of Gauss’s Law.

• Gauss’s Law (Electric flux through a closed surface)

  • Gauss’s Law relates:
    • electric charge distribution (enclosed charge)
    • to the electric field created.
  • The number of electric field lines (electric flux) passing through a closed surface is proportional to the enclosed charge, divided by the permittivity of the surrounding medium (air/vacuum model).

• Key equation for electric flux (as used in the video)

  • Electric flux through a surface: [ \Phi = EA\cos\theta ]

  • Where:

    • (E) = electric field strength (unit: N/C)
    • (A) = area of the surface (m²)
    • (\theta) = angle between the electric field direction and the surface normal
  • The video explains three special geometric cases based on (\theta).

• Three angle/field-line cases for flux

  1. Field direction parallel to the plane
     • (\theta = 90^\circ)
     • (\cos 90^\circ = 0)
     • (\Rightarrow \Phi = 0)
  2. Field direction perpendicular to the plane
     • (\theta = 0^\circ)
     • (\cos 0^\circ = 1)
     • (\Rightarrow \Phi = EA)
  3. Field not perpendicular to the plane
     • Use the general form: [ \Phi = EA\cos\theta ]

• Electric potential energy

  • Defined as the work done by the Coulomb force to move a test charge from one point to another (around the source charge).

  • Formula: [ E_p = k\frac{Q_1Q_2}{r} ]

  • Notes:

    • Scalar quantity → must include charge signs
    • Units: Joule

• Electric potential (potential difference idea)

  • Electric potential is potential energy per unit charge: [ V=\frac{E_p}{Q_2} = k\frac{Q_1}{r} ]

  • Also a scalar quantity → include charge signs

  • With multiple source charges, potentials add: [ V_{total}=V_1+V_2+\cdots+V_n ]

• Relationship between work and electric potential

  • Work relates to change in electric potential energy: [ W=\Delta E_p = E_{p2}-E_{p1} ]

  • Substituting yields a common form: [ W = q\Delta V = q\,(V_2 - V_1) ]

  • Units: Joule

• Conservation of mechanical energy in an electric field

  • Mechanical energy is conserved for charged particle motion under electrostatic forces: [ E_{m1}=E_{m2} ]

  • Expanded as: [ qV_1+\frac{1}{2}mv_1^2 = qV_2+\frac{1}{2}mv_2^2 ]

  • Used later to solve for final velocity.

• Static electricity formulas recalled

  • Coulomb force: [ F = k\frac{Q_1Q_2}{r^2} ]

  • Electric field: [ E = k\frac{Q}{r^2} ]

  • Electric potential energy: [ E_p = k\frac{q}{r} ]

  • Electric potential: [ V = k\frac{Q}{r} ]

---

Method / instructions used for solving example problems

1) Electric flux through an equilateral triangle in a uniform field

• Given

  • Side length (s = 20\sqrt{3}\,\text{cm})
  • Uniform electric field magnitude (E = 240\,\text{N/C})
  • Field makes specific angles with the triangle plane.

• Steps

  • Find triangle height (equilateral triangle geometry):
    • Use Pythagorean relation for equilateral triangle to compute (h) from half-side and side geometry.

  • Compute area: [ A=\frac{(base)(height)}{2} ]

    • Convert area to m².
      • Use flux formula: [ \Phi = EA\cos\theta ]

  • Apply angle cases:

    • If field is parallel to plane → (\theta=90^\circ) → (\Phi=0)
    • If field is perpendicular to plane → (\theta=0^\circ) → (\Phi=EA)
    • If field makes 53° with plane:
      • Convert to angle with normal:
        • (\theta = 37^\circ) (normal is (90^\circ) to the plane)
      • Compute: [ \Phi=EA\cos(37^\circ) ]

2) Gauss’s law style: determine enclosed charge from flux through a square

• Given

  • Electric field (E = 4000\,\text{N/C} = 4\times 10^3)
  • Square side (s = 10\,\text{cm}) → area (A = s^2)
  • Angle (\theta = 60^\circ)
  • Permittivity (\varepsilon_0 = 8.85\times 10^{-12})

• Steps

  • Start from Gauss-related flux relation: [ \Phi = \frac{Q_{enclosed}}{\varepsilon_0} ]

  • Using: [ \Phi = EA\cos\theta ]

  • Solve for enclosed charge: [ Q=\varepsilon_0\frac{EA\cos\theta}{1} ]

  • Substitute and compute (Q).

3) Determine source charge from electric potential energy

• Given

  • Distance (R = 3\times 10^{-4}\,\text{m})
  • Test charge (Q_2 = -6\times 10^{-7}\,\text{C})
  • Potential energy (E_p = 18\,\text{J})

• Steps

  • Use: [ E_p = k\frac{Q_1Q_2}{R} ]

  • Rearrange to solve for (Q_1): [ Q_1 = \frac{E_pR}{kQ_2} ]

  • Emphasize sign handling because (E_p) is scalar but depends on charge signs.

4) Electric potential at center of a rectangle from multiple charges

• Given

  • Rectangle dimensions: length 80 cm, width 60 cm
  • Four corner charges:
    • (Q_1=10\,\mu\text{C})
    • (Q_2=20\,\mu\text{C})
    • (Q_3=-30\,\mu\text{C})
    • (Q_4=40\,\mu\text{C})

• Steps

  • Find rectangle diagonal: [ AC=\sqrt{80^2+60^2}=100\,\text{cm} ]

  • Distance from center to each corner is half the diagonal: [ R=\frac{AC}{2}=50\,\text{cm}=0.5\,\text{m} ]

  • Compute total potential using superposition: [ V_{total}=k\left(\frac{Q_1}{R}+\frac{Q_2}{R}+\frac{Q_3}{R}+\frac{Q_4}{R}\right) ]

  • Include charge signs in each (Q_i).

  • Convert units if needed (video expresses result as 720 kV).

5) Work needed using charge and two potentials

• Given

  • Charge (q=30\,\text{C})
  • (V_1 = 2\times 10^6\,\text{V})
  • (V_2 = 1.2\times 10^7\,\text{V} = 12\times 10^6\,\text{V})

• Steps

  • Use: [ W=q\,(V_2-V_1)=q\Delta V ]

  • Substitute to compute (W).

6) Final electron speed using energy conservation with potential difference

• Given

  • Electron mass (m = 9\times 10^{-31}\,\text{kg})
  • Electron charge (q = -1.6\times 10^{-19}\,\text{C})
  • Initial velocity (v_1=0)
  • Potential difference (\Delta V = 4500\,\text{V} = 4.5\times 10^3\,\text{V})

• Steps

  • Apply conservation of mechanical energy in electric field: [ qV_1+\frac12 mv_1^2=qV_2+\frac12 mv_2^2 ]

  • Rearrange using (\Delta V = V_2 - V_1): [ \frac12 mv_1^2-\frac12 mv_2^2=q\Delta V ]

  • With (v_1=0), solve for (v_2).

  • Substitute numerical values and compute (v_2).

---

Speakers / sources featured

• Gia Academy YouTube channel — presenter/instructor and problem-solving narration
• Carl Friedrich Gauss — historically referenced mathematician and physicist