Class 12 Physics All Formulas | CBSE, ICSE, State Boards

Complete Class 12 Physics Formulas

Comprehensive reference with examples, calculators, and board-specific content for CBSE, ICSE & State Boards. Essential for JEE, NEET preparation.

Electrostatics

Coulomb’s Law
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F = k(q₁q₂)/r²
The electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them.

Variables:

  • F Force (N)
  • k Coulomb’s Constant (9×10⁹ N·m²/C²)
  • q₁, q₂ Charges (C)
  • r Distance (m)

Example

Two charges of 2μC and 3μC are 0.1 m apart. Force between them is:

F = (9×10⁹ × 2×10⁻⁶ × 3×10⁻⁶) / (0.1)² = 5.4 N

Calculate Force

Electric Field
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E = kQ/r²
Electric field intensity at a point is the force experienced per unit positive charge placed at that point.

Variables:

  • E Electric Field (N/C)
  • k Coulomb’s Constant
  • Q Source Charge (C)
  • r Distance (m)

Example

Electric field at 0.2 m from 5μC charge:

E = (9×10⁹ × 5×10⁻⁶) / (0.2)² = 1.125×10⁶ N/C

Calculate Electric Field

Electric Potential
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V = kQ/r
Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point.

Variables:

  • V Electric Potential (V)
  • k Coulomb’s Constant
  • Q Charge (C)
  • r Distance (m)

Example

Potential at 0.1 m from 2μC charge:

V = (9×10⁹ × 2×10⁻⁶) / 0.1 = 1.8×10⁵ V

Calculate Potential

Capacitance
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C = Q/V
Capacitance is the ability of a conductor to store charge per unit potential difference.

Variables:

  • C Capacitance (F)
  • Q Charge (C)
  • V Potential (V)

Example

A capacitor stores 10μC charge at 100V. Its capacitance is:

C = 10×10⁻⁶ / 100 = 0.1μF

Calculate Capacitance

Current Electricity

Ohm’s Law
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V = IR
The potential difference across a conductor is directly proportional to the current flowing through it, provided temperature remains constant.

Variables:

  • V Voltage (V)
  • I Current (A)
  • R Resistance (Ω)

Example

If 2A current flows through 10Ω resistor, voltage is:

V = 2 × 10 = 20V

Calculate Voltage

Resistance
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R = ρL/A
The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area.

Variables:

  • R Resistance (Ω)
  • ρ Resistivity (Ω·m)
  • L Length (m)
  • A Area (m²)

Example

A copper wire (ρ=1.7×10⁻⁸ Ω·m) of length 10m and area 1mm² has resistance:

R = (1.7×10⁻⁸ × 10) / (1×10⁻⁶) = 0.17Ω

Calculate Resistance

Power in Circuit
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P = VI = I²R = V²/R
Electrical power is the rate at which electrical energy is consumed in a circuit.

Variables:

  • P Power (W)
  • V Voltage (V)
  • I Current (A)
  • R Resistance (Ω)

Example

A 100W bulb at 220V draws current:

I = P/V = 100/220 = 0.455A

Calculate Power

Electromagnetic Waves

Wave Equation
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c = fλ
The relationship between speed, frequency, and wavelength of electromagnetic waves.

Variables:

  • c Speed (3×10⁸ m/s)
  • f Frequency (Hz)
  • λ Wavelength (m)

Example

Radio wave with frequency 100 MHz has wavelength:

λ = c/f = 3×10⁸/100×10⁶ = 3 m

Calculate Wavelength

Energy of Photon
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E = hf
The energy of a photon is directly proportional to its frequency.

Variables:

  • E Energy (J)
  • h Planck’s Constant (6.63×10⁻³⁴ J·s)
  • f Frequency (Hz)

Example

Photon of yellow light (λ=600nm) has energy:

f = c/λ = 5×10¹⁴ Hz

E = 6.63×10⁻³⁴ × 5×10¹⁴ = 3.315×10⁻¹⁹ J

Calculate Photon Energy

Wave Optics

Young’s Double Slit
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β = λD/d
The fringe width in Young’s double slit experiment is directly proportional to wavelength and screen distance, and inversely proportional to slit separation.

Variables:

  • β Fringe Width (m)
  • λ Wavelength (m)
  • D Screen Distance (m)
  • d Slit Separation (m)

Example

For λ=600nm, D=1m, d=1mm, fringe width is:

β = (600×10⁻⁹ × 1)/(1×10⁻³) = 0.6 mm

Calculate Fringe Width

Dual Nature of Radiation & Matter

Photoelectric Effect
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Kmax = hf – φ
Einstein’s photoelectric equation: Maximum kinetic energy of photoelectrons equals photon energy minus work function.

Variables:

  • Kmax Max KE (J)
  • h Planck’s Constant
  • f Frequency (Hz)
  • φ Work Function (J)

Example

For metal with φ=2eV, light of λ=300nm:

f = c/λ = 1×10¹⁵ Hz

Kmax = (6.63×10⁻³⁴×1×10¹⁵) – (2×1.6×10⁻¹⁹) = 2.63×10⁻¹⁹ J

Calculate Max KE

de Broglie Wavelength
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λ = h/p
Every moving particle has wave characteristics with wavelength inversely proportional to its momentum.

Variables:

  • λ Wavelength (m)
  • h Planck’s Constant
  • p Momentum (kg·m/s)

Example

Electron with energy 100eV has wavelength:

p = √(2mE) = √(2×9.1×10⁻³¹×100×1.6×10⁻¹⁹)

λ = 6.63×10⁻³⁴/p = 1.23×10⁻¹⁰ m

Calculate de Broglie Wavelength

Atoms & Nuclei

Energy Levels
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En = -13.6/n² eV
Energy of electron in nth orbit of hydrogen atom. Negative sign indicates bound state.

Variables:

  • En Energy of nth level (eV)
  • n Quantum Number

Example

Energy of electron in ground state (n=1):

E₁ = -13.6 eV

Energy emitted in transition from n=3 to n=2:

ΔE = 13.6(1/4 – 1/9) = 1.89 eV

Calculate Energy Level

Radioactive Decay
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N = N₀e^(-λt)
The number of radioactive nuclei decreases exponentially with time according to the law of radioactive decay.

Variables:

  • N Nuclei at time t
  • N₀ Initial nuclei
  • λ Decay Constant
  • t Time

Example

Half-life T½ = ln2/λ = 0.693/λ

For carbon-14, T½=5730 years, λ=3.84×10⁻¹² s⁻¹

Calculate Remaining Nuclei

Electronic Devices

Transistor Current
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IE = IB + IC
In a transistor, emitter current equals the sum of base current and collector current.

Variables:

  • IE Emitter Current
  • IB Base Current
  • IC Collector Current

Example

For transistor with β=100, if IB=10μA:

IC = βIB = 1mA

IE = IC + IB = 1.01mA

Calculate Transistor Currents

Communication Systems

Amplitude Modulation
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Am(t) = Ac[1 + macos(ωmt)]cos(ωct)
In AM, the amplitude of carrier wave is varied in accordance with the amplitude of modulating signal.

Variables:

  • Am Modulated Amplitude
  • Ac Carrier Amplitude
  • ma Modulation Index
  • ωm Modulating Frequency

Example

For ma = 0.8, sideband frequencies are fc ± fm

Bandwidth = 2fm

Calculate AM Bandwidth

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