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CLASS 12 • PHYSICS • CHAPTER 1

Electric Charges and Fields

High-yield revision notes, key concepts, formulas, and CBSE board examination questions for Class 12 Physics.

Coulomb's Law & Vector Formulation

The electrostatic force between two stationary point charges $q_1$ and $q_2$ in vacuum separated by distance $r$:

$$\vec{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{21}$$

Where $\varepsilon_0 = 8.854 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$ is vacuum permittivity, and $\frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$.


Electric Dipole & Torque

An electric dipole consists of two equal and opposite charges $\pm q$ separated by distance $2a$. Dipole moment vector: $\vec{p} = q(2a)\hat{p}$ (directed from $-q$ to $+q$).

  1. Electric Field on Axial Line ($r \gg a$):

$$E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3}$$

  1. Electric Field on Equatorial Line ($r \gg a$):

$$E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} \implies E_{\text{axial}} = 2 E_{\text{equatorial}}$$

  1. Torque in Uniform Electric Field $\vec{E}$:

$$\vec{\tau} = \vec{p} \times \vec{E} \implies \tau = pE\sin\theta$$


Gauss's Law & Applications

The total electric flux passing through any closed Gaussian surface equals $\frac{1}{\varepsilon_0}$ times the net charge enclosed:

$$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$

Derivations:

  1. Infinitely Long Straight Uniformly Charged Wire ($\lambda$):

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

  1. Uniformly Charged Infinite Thin Sheet ($\sigma$):

$$E = \frac{\sigma}{2\varepsilon_0} \quad (\text{independent of distance } r!)$$

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