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01Class 12 Physics

Electric Charges and Fields formulas

Coulomb's law, electric field and Gauss's law formulas for Electric Charges and Fields. Symmetry decides which Gauss's law result to use.

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Charge and force

Quantisation of charge
q=ne\displaystyle q = ne

e = 1.6 × 10⁻¹⁹ C.

Coulomb's law
F=14πε0q1q2r2\displaystyle F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}

1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻².

Electric field of a point charge
E=14πε0qr2\displaystyle E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}
Force on a charge in a field
F⃗=qE⃗\displaystyle \vec F = q\vec E

Electric dipole

Dipole moment
p⃗=q(2a⃗)\displaystyle \vec p = q(2\vec a)

Points from negative to positive charge.

Field on the axis
E=14πε02pr3\displaystyle E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}

For r much larger than a.

Field on the equatorial line
E=14πε0pr3\displaystyle E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}

Opposite in direction to p.

Torque in a uniform field
τ⃗=p⃗×E⃗\displaystyle \vec\tau = \vec p\times\vec E

Gauss's law

Electric flux
Φ=∮E⃗⋅dA⃗=qenclosedε0\displaystyle \Phi = \oint\vec E\cdot d\vec A = \frac{q_{\text{enclosed}}}{\varepsilon_0}
Infinite line of charge
E=λ2πε0r\displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 r}
Infinite plane sheet
E=σ2ε0\displaystyle E = \frac{\sigma}{2\varepsilon_0}
Outside a charged spherical shell
E=14πε0qr2\displaystyle E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}

Inside the shell, E = 0.

Where students slip up

  • Using the plane-sheet result next to a conductor. Just outside a charged conductor, E = σ/ε₀.
  • Forgetting flux depends only on the enclosed charge, while the field at a point depends on all charges.
  • Adding field magnitudes directly. Electric fields add as vectors.

Try it on your weakest chapter.

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