Class XII · Physics · Chapter 1 · Electrostatics

Electrostatics
Derivations

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Akshit Tyagi
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Akshit Tyagi
Class XII · Physics Fair Notes
01 · Coulomb's Law 02 · Dipole Field 03 · Torque 04 · Gauss's Theorem 05 · Applications
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01
Coulomb's Law — Vector Form
Force between two point charges
The force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them — acting along the line joining them.
+q₁ +q₂ F⃗₁₂ F⃗₂₁ r₁₂ r̂₁₂ (unit vector from q₁ to q₂) r⃗₁ r⃗₂

Fig 1.1 — Like charges repel; F₁₂ = force on q₂ due to q₁

Derivation

Let charges $q_1$ and $q_2$ be at positions $\vec{r}_1$ and $\vec{r}_2$. The vector pointing from $q_1$ to $q_2$ is:

$$\vec{r}_{12} = \vec{r}_2 - \vec{r}_1, \qquad |\vec{r}_{12}| = r$$

The unit vector $\hat{r}_{12}$ from $q_1$ toward $q_2$:

$$\hat{r}_{12} = \frac{\vec{r}_{12}}{r} = \frac{\vec{r}_2-\vec{r}_1}{|\vec{r}_2-\vec{r}_1|}$$
①
By Coulomb's law the magnitude of force is $F = k\dfrac{|q_1||q_2|}{r^2}$, where $k = \dfrac{1}{4\pi\varepsilon_0} \approx 9\times10^9\;\text{N\,m}^2\text{C}^{-2}$.
②
In vector form, the force on $q_2$ due to $q_1$ (along $\hat{r}_{12}$): $$\vec{F}_{12} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\,\hat{r}_{12}$$
③
Substituting $\hat{r}_{12} = \dfrac{\vec{r}_2-\vec{r}_1}{r}$:
★ Result $$\vec{F}_{12} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^3}\,(\vec{r}_2-\vec{r}_1)$$
Newton's 3rd Law: $\vec{F}_{21} = -\vec{F}_{12}$ always.  |  Sign rule: $q_1 q_2 > 0$ → repulsive; $q_1 q_2 < 0$ → attractive.
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02
Electric Field due to an Electric Dipole
Axial Point & Equatorial Point
An electric dipole is a pair of equal and opposite charges ($+q$ and $-q$) separated by distance $2a$. Dipole moment: $\vec{p} = q\cdot2a$, directed from $-q$ to $+q$.
Case I — Axial Point (End-on)
axis O −q +q P a a r (from centre) E₊ E₋ p⃗ (dipole moment)

Fig 2.1 — Point P on axial line, distance r from centre O

Derivation — Axial

Distance from $+q$ to P = $(r-a)$; from $-q$ to P = $(r+a)$.

①
Field at P due to $+q$ (along $+\hat{p}$): $\displaystyle E_+ = \frac{q}{4\pi\varepsilon_0(r-a)^2}$
②
Field at P due to $-q$ (along $-\hat{p}$): $\displaystyle E_- = \frac{q}{4\pi\varepsilon_0(r+a)^2}$
③
Net field $E_{ax} = E_+ - E_-$: $$\frac{q}{4\pi\varepsilon_0}\!\left[\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\right] =\frac{q}{4\pi\varepsilon_0}\cdot\frac{4ar}{(r^2-a^2)^2}$$
★ Result $$\vec{E}_{axial} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2pr}{(r^2-a^2)^2}\;\hat{p}$$
For $r\gg a$:  $E_{axial} \approx \dfrac{2p}{4\pi\varepsilon_0 r^3}$
Case II — Equatorial Point (Broad-side on)
O −q +q P E₊ E₋ E⃗_eq r a p⃗ (antiparallel to p⃗)

Fig 2.2 — Point P on perpendicular bisector. ⊥ components cancel; axial components add.

Derivation — Equatorial

Distance from either charge to P: $r_1 = \sqrt{r^2+a^2}$

①
By symmetry, $|E_+| = |E_-| = \dfrac{q}{4\pi\varepsilon_0(r^2+a^2)}$
②
Components perpendicular to dipole axis cancel. Components along the axis (opposite to $\vec{p}$) add up. Each has a factor $\cos\theta = \dfrac{a}{\sqrt{r^2+a^2}}$.
③
$$E_{eq} = 2E_+\cdot\frac{a}{\sqrt{r^2+a^2}} = \frac{2qa}{4\pi\varepsilon_0(r^2+a^2)^{3/2}}$$
★ Result $$\vec{E}_{eq} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{p}{(r^2+a^2)^{3/2}}\;\;(-\hat{p})$$
For $r\gg a$:  $E_{eq} \approx \dfrac{p}{4\pi\varepsilon_0 r^3}$  — exactly half of $E_{axial}$. Direction is opposite to $\vec{p}$.
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03
Torque on an Electric Dipole
In a Uniform Electric Field
In a uniform electric field, the net force on a dipole is zero (equal and opposite forces). But the forces form a couple which produces a torque that tends to align the dipole along $\vec{E}$.
E⃗ −q +q qE⃗ qE⃗ p⃗ θ d = 2a sinθ 2a τ

Fig 3.1 — Dipole at angle θ in uniform E; forces form a couple of torque τ = pE sinθ

Derivation
①
Force on $+q$: $\vec{F}_+ = q\vec{E}$ (along $\vec{E}$).    Force on $-q$: $\vec{F}_- = -q\vec{E}$ (opposite to $\vec{E}$).
②
Net force $= q\vec{E} - q\vec{E} = \vec{0}$  →  No translation. But the two forces form a couple → rotational effect (torque).
③
Torque = Force × perpendicular distance between forces: $$\tau = qE \times d = qE \times 2a\sin\theta$$
④
Since $p = q \times 2a$ (dipole moment): $$\tau = pE\sin\theta$$
★ Result $$\vec{\tau} = \vec{p}\times\vec{E}$$ $$\text{Magnitude: }\tau = pE\sin\theta$$
Stable eq. ($\theta=0°$): $\tau=0$, dipole aligned with $\vec{E}$.  |  Unstable eq. ($\theta=180°$): $\tau=0$, antiparallel to $\vec{E}$.
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04
Gauss's Theorem — with Proof
Electric flux · Closed surface · Enclosed charge
Statement: The total electric flux through any closed surface is equal to the net charge enclosed inside it divided by $\varepsilon_0$, regardless of the shape of the surface.
★ Gauss's Law $$\oint_S \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$$
+q r dA⃗ Gaussian Surface (S) E⃗ ∥ dA⃗ everywhere on sphere → E·dA = E dA

Fig 4.1 — Spherical Gaussian surface of radius r centred on +q

Proof (Point Charge)
①
Consider point charge $+q$ at origin. Choose a Gaussian sphere of radius $r$. Electric field at any surface point: $E = \dfrac{q}{4\pi\varepsilon_0 r^2}$ (radially outward).
②
At every point on the sphere, $\vec{E}$ is radially outward and $d\vec{A}$ is also radially outward. So $\vec{E}\cdot d\vec{A} = E\,dA$ (angle between them = 0°).
③
Total flux: $$\Phi = \oint \vec{E}\cdot d\vec{A} = E\oint dA = \frac{q}{4\pi\varepsilon_0 r^2}\times 4\pi r^2$$
★ Proved $$\Phi = \oint \vec{E}\cdot d\vec{A} = \frac{q}{\varepsilon_0}$$
Key insight: Result is independent of radius $r$. The theorem holds for any closed surface of any shape — the Gaussian surface is just a mathematical tool.
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05
Applications of Gauss's Theorem
Line Charge · Plane Sheet · Spherical Shell
Gauss's theorem is most powerful when charge distribution has symmetry. We choose a Gaussian surface matching that symmetry so $\vec{E}$ is constant and parallel to $d\vec{A}$ — making the surface integral trivial.
① Infinite Line Charge
λ C/m Gaussian cylinder E⃗ r L

Gaussian surface: Cylinder of radius $r$, length $L$, coaxial with line charge $\lambda$.

Flux through flat ends = 0 ($\vec{E}\perp d\vec{A}$). Flux through curved surface:

$$E\cdot 2\pi rL = \frac{\lambda L}{\varepsilon_0}$$
★ $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$
Note: $E \propto \dfrac{1}{r}$ for a line charge.
② Infinite Plane Sheet
σ C/m² E⃗ E⃗ A A Pillbox

Gaussian surface: Pillbox (cylinder) with flat caps of area $A$ parallel to sheet.

Flux through curved surface = 0. Flux through both caps:

$$2E\cdot A = \frac{\sigma A}{\varepsilon_0}$$
★ $$E = \frac{\sigma}{2\varepsilon_0}$$
Note: $E$ is uniform and independent of distance from sheet.
③ Uniformly Charged Spherical Shell
Shell (total charge Q) E⃗ E = 0 (inside shell) R r

Fig 5.3 — Inner & outer Gaussian surfaces for shell

r E O R E = kQ/r² E = 0 kQ/R² E vs r (spherical shell)

Fig 5.4 — E jumps at r = R, then falls as 1/r²

Derivation
Case 1: Outside shell ($r > R$)

Gaussian sphere of radius $r$ encloses total charge $Q$.

$$E\cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}$$
★ $$E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2}$$ Acts like a point charge
Case 2: Inside shell ($r < R$)

Gaussian sphere of radius $r$ encloses no charge ($q_{enc}=0$).

$$\oint\vec{E}\cdot d\vec{A} = \frac{0}{\varepsilon_0} = 0$$
★ $$E = 0 \text{ inside the shell}$$ Remarkable result!
At the surface ($r = R$):  $E_{max} = \dfrac{Q}{4\pi\varepsilon_0 R^2}$   — highest field value. The field is discontinuous at the surface.