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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 01 (Electrostatics)

Chapter 02: Electrostatic Potential and Capacitance

Comprehensive Revision Dossier • High-Yield Formulas, Derivation Blueprints & Exam Trap Sheet

1. Definitive Formula & Dimensions Matrix

Concept Mathematical Formula SI Unit / Dim Boundary Conditions & Notes
Electric Potential ($V$)[cite: 13] $$V = -\int_\infty^r \vec{E}\cdot d\vec{r} = \frac{W^{\text{ext}}}{q_0}\text{[cite: 13]}$$ $\text{Volt (V)} = \text{J/C}$
$[\text{ML}^2\text{T}^{-3}\text{A}^{-1}]$		 Strict scalar quantity[cite: 13]; charges substituted with algebraic signs[cite: 13]. Reference $V(\infty) = 0$[cite: 13].
Point Charge Field[cite: 13] $$V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}\text{[cite: 13]}$$ $\text{V}$ Potential decreases as $1/r$[cite: 13]; concentric equipotential spheres[cite: 13].
Short Dipole ($r \gg a$)[cite: 13] $$V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}\text{[cite: 13]}$$ $\text{V}$ Axial: $\pm\frac{kp}{r^2}$[cite: 13]; Equatorial ($\theta=90^\circ$): $V_{\text{eq}} = 0$ at all distances[cite: 13].
Potential Gradient[cite: 13] $$\vec{E} = -\vec{\nabla}V \implies E = -\frac{dV}{dr}\text{[cite: 13]}$$ $\text{V/m} = \text{N/C}$ Electric field points in the direction where potential drops steepest[cite: 13].
System PE (2 Charges)[cite: 13] $$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}\text{[cite: 13]}$$ $\text{Joule (J)}$ For $N$ charges, sum over all $\frac{N(N-1)}{2}$ unique pairs[cite: 13]. Like charges $\implies U > 0$[cite: 13].
Parallel Plate ($C_0$)[cite: 13] $$C_0 = \frac{\varepsilon_0 A}{d}\text{[cite: 13]}$$ $\text{Farad (F)} = \text{C/V}$ Pure geometric quantity; completely independent of applied $V$ or $Q$[cite: 13].
Partial Dielectric Slab[cite: 13] $$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}\text{[cite: 13]}$$ $\text{F}$ Full insertion ($t=d$): $C = KC_0$[cite: 13]. Metal slab ($K \to \infty$): $C = \frac{\varepsilon_0 A}{d-t}$.
Series Network[cite: 13] $$\frac{1}{C_s} = \sum_{i=1}^n \frac{1}{C_i}\text{[cite: 13]}$$ $\text{F}$ Charge $Q$ is identical across each capacitor[cite: 13]; equivalent $C_s < C_{\min}$[cite: 13].
Parallel Network[cite: 13] $$C_p = \sum_{i=1}^n C_i\text{[cite: 13]}$$ $\text{F}$ Voltage $V$ is identical across all branches[cite: 13]; equivalent $C_p > C_{\max}$[cite: 13].
Stored Energy ($U$)[cite: 13] $$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}\text{[cite: 13]}$$ $\text{J}$ Stored in the electrostatic field volume[cite: 13]. Energy density $u_E = \frac{1}{2}\varepsilon_0 E^2$.
Charge Sharing Loss $$\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$$ $\text{J}$ Always $\ge 0$; dissipated as Joule heat in connecting wires and EM radiation.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)[cite: 13]

Blueprint 2.1: Relation between Electric Field & Potential Gradient ($E = -\frac{dV}{dr}$)[cite: 13]

1. Consider two closely spaced equipotential surfaces $A$ and $B$ separated by perpendicular distance $dr$, having potentials $V$ and $V - dV$[cite: 13].

2. Move a unit positive charge ($q_0 = +1\text{ C}$) from surface $B$ to surface $A$ against the electric field $\vec{E}$[cite: 13]. External force required: $\vec{F}_{\text{ext}} = -\vec{E}$.

3. Work done by external agent across displacement $dr$[cite: 13]:

$$dW_{\text{ext}} = \vec{F}_{\text{ext}}\cdot d\vec{r} = -E\,dr$$

4. By fundamental definition, work done per unit charge equals potential difference $V_A - V_B$[cite: 13]:

$$V_A - V_B = V - (V - dV) = dV$$

5. Equating work done to potential difference:

$$-E\,dr = dV \implies E = -\frac{dV}{dr}\text{[cite: 13]}$$

Blueprint 2.2: Electric Potential due to a Dipole at Arbitrary Coordinates $(r, \theta)$[cite: 13]

1. Consider a dipole consisting of $-q$ and $+q$ separated by $2a$[cite: 13]. Point $P$ is at distance $r$ from midpoint $O$, making an angle $\theta$ with $\vec{p}$[cite: 13].

2. For a short dipole ($r \gg a$), path distances from individual charges are:

$$r_1 \approx r - a\cos\theta \quad (\text{from }+q), \qquad r_2 \approx r + a\cos\theta \quad (\text{from }-q)$$

3. Total scalar potential is the algebraic sum $V = V_+ + V_-$[cite: 13]:

$$V = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r_1} - \frac{1}{r_2}\right] = \frac{q}{4\pi\varepsilon_0}\left[\frac{(r + a\cos\theta) - (r - a\cos\theta)}{r^2 - a^2\cos^2\theta}\right] \approx \frac{q(2a\cos\theta)}{4\pi\varepsilon_0 r^2}$$

4. Substitute dipole moment $p = q(2a)$[cite: 13]:

$$V(r, \theta) = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{\vec{p}\cdot\hat{r}}{r^2} \quad \left(V_{\text{axial}} = \pm\frac{kp}{r^2}, \quad V_{\text{eq}} = 0\right)\text{[cite: 13]}$$

Blueprint 2.3: Parallel Plate Capacitor with Partial Dielectric Slab of Thickness $t < d$[cite: 13]

1. Parallel plates have area $A$, separation $d$, and free surface charge density $\sigma = \frac{Q}{A}$[cite: 13]. Vacuum field is $E_0 = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$[cite: 13].

2. When a dielectric slab of thickness $t$ and constant $K$ is inserted, the field in the slab drops to $E = \frac{E_0}{K}$[cite: 13], while the remaining vacuum gap of thickness $(d - t)$ maintains field $E_0$[cite: 13].

3. Total potential difference across the plates is the sum of potentials in both zones[cite: 13]:

$$V = E_0(d - t) + E t = E_0(d - t) + \frac{E_0}{K}t = E_0\left[(d - t) + \frac{t}{K}\right]$$

4. Substitute $E_0 = \frac{Q}{\varepsilon_0 A}$ into $C = \frac{Q}{V}$[cite: 13]:

$$C = \frac{Q}{\frac{Q}{\varepsilon_0 A}\left[d - t + \frac{t}{K}\right]} \implies C = \frac{\varepsilon_0 A}{d - t\left(1 - \frac{1}{K}\right)}\text{[cite: 13]}$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap: Insertion of Dielectric Slab ($K$) under Different Boundary States[cite: 13]

Physical Parameter Case A: Battery Remains Connected Case B: Battery Disconnected
Capacitance ($C$) Increases to $K C_0$[cite: 13] Increases to $K C_0$[cite: 13]
Potential Difference ($V$) CONSTANT ($V = V_0$) Decreases to $V_0 / K$[cite: 13]
Charge on Plates ($Q$) Increases to $K Q_0$ CONSTANT ($Q = Q_0$)
Electric Field ($E$) CONSTANT ($E = V_0/d$) Decreases to $E_0 / K$[cite: 13]
Stored Energy ($U$) Increases to $K U_0$ Decreases to $U_0 / K$[cite: 13]

⚠️ Trap: Work Done Semantics

$$W_{\text{ext}} = q\Delta V = +q(V_B - V_A)$$ $$W_{\text{field}} = -\Delta U = -q(V_B - V_A)$$ Notice the opposing negative sign! Work done *by the field* reduces potential energy, whereas work done *by an external agent* increases it.

⚠️ Trap: Equipotential Field Orthogonality

Because $W = q(V_B - V_A) = 0$ along any path on an equipotential surface[cite: 13]: $$\int \vec{E}\cdot d\vec{r} = 0 \implies \vec{E} \perp \text{Surface everywhere}\text{[cite: 13]}$$ If $\vec{E}$ had any parallel tangential component, moving a charge would require non-zero work.

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Multi-Charge Square Lattice Binding Energy Exemplar 01

Four charges $+q, -q, +q, -q$ are placed at vertices $A, B, C, D$ of a square of side $d$ taken in order. Find the potential energy of the configuration, and determine the work required to disassemble it to infinity.

Solution: Total pairs for $N=4$ is $\frac{4(3)}{2} = 6$ unique pairs.

• 4 adjacent side pairs (separation $d$): $(+q)(-q) = -q^2$ each.

• 2 diagonal pairs (separation $d\sqrt{2}$): $(+q)(+q) = +q^2$ and $(-q)(-q) = +q^2$.

$$U = \frac{1}{4\pi\varepsilon_0}\left[ 4\left(\frac{-q^2}{d}\right) + 2\left(\frac{q^2}{d\sqrt{2}}\right) \right] = -\frac{q^2}{4\pi\varepsilon_0 d}\left(4 - \sqrt{2}\right)$$

To disassemble to infinity ($U_f = 0$):

$$W_{\text{ext}} = U_f - U_i = +\frac{q^2}{4\pi\varepsilon_0 d}\left(4 - \sqrt{2}\right)\text{ Joules}$$
Problem 2: Capacitor Charge Sharing & Energy Dissipation Exemplar 02

A $600\text{ pF}$ capacitor charged to $200\text{ V}$ is disconnected from its supply and connected in parallel to an uncharged $600\text{ pF}$ capacitor[cite: 13]. Calculate the total energy lost in the process[cite: 13].

Solution:

1. Initial stored energy $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2}(600 \times 10^{-12})(200)^2 = 1.2 \times 10^{-5}\text{ J}$[cite: 13].

2. When connected in parallel, common potential $V_c = \frac{C_1 V_1 + 0}{C_1 + C_2} = \frac{600 \times 200}{1200} = 100\text{ V}$[cite: 13].

3. Final total energy $U_f = \frac{1}{2}(C_1 + C_2)V_c^2 = \frac{1}{2}(1200 \times 10^{-12})(100)^2 = 0.6 \times 10^{-5}\text{ J}$[cite: 13].

$$\Delta U = U_i - U_f = 0.6 \times 10^{-5}\text{ J} = 6\text{ }\mu\text{J}\quad (\text{50% energy lost as heat and sparks})$$

5. Printable Pocket Flashcard Grid

Equipotential Work

$W = q\Delta V = 0$[cite: 13]

Gradient Law

$E = -\frac{dV}{dr}$[cite: 13]

Partial Slab

$C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$[cite: 13]

Energy Density

$u_E = \frac{1}{2}\varepsilon_0 E^2$