Axiom Physics | Class 12 - Electric Potential & Potential Difference

⚡ 01. The Spark (Analogy)

Socho ek rolling ball ko surface par chhod diya jaye: woh naturally high altitude se low altitude ki taraf roll karegi bina kisi fuel ke. Gravitational landscape mein jaise "height" tai karti hai ki cheez kahan giregi, theek waise hi electrostatics mein Electric Potential ($V$) ek "electrical altitude" hai. Ek positive charge hamesha High Potential se Low Potential ki taraf bhagta hai, jabki ek negative charge upar ki taraf (Low to High Potential) climb karta hai.

📐 02. The Core Law & Formulae

Electric Potential ($V$) & Difference ($\Delta V$)

External work required per unit test charge without acceleration from infinity:

$$V = \frac{W_{\infty \to r}^{\text{ext}}}{q_0} = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}, \quad \Delta V = V_B - V_A = \frac{W_{A \to B}^{\text{ext}}}{q_0}$$

SI Unit: $\text{Volt } (\text{V}) = \text{Joule / Coulomb } (\text{J/C})$. Scalar quantity with sign.

Point Charge & Short Dipole Potential

Radial falloff comparisons:

$$V_{\text{point}} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}, \quad V_{\text{dipole}} = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$$

Note: Dipole potential on equatorial axis ($\theta = 90^\circ$) is identically zero at all distances[cite: 13]!

03. Interactive Visualizer: Potential Contour & Test-Charge Drift Engine

Source Charge Magnitude ($Q$) +5 nC

Probe Distance ($r$) 10.0 cm

Probe Particle Sign +q₀ (Rolls Down)

Potential at Probe ($V = kQ/r$):

+449.5 V

Field Strength ($E = -dV/dr$):

4.50 × 10³ V/m

Potential Energy ($U = q_0 V$):

+4.50 × 10⁻⁷ J

📝 04. Stepwise Derivations Blueprint[cite: 13]

Derivation A

Electric Potential due to an Isolated Point Charge $q$[cite: 13]

1. Place a source charge $+q$ at the origin $O$. We wish to find potential at point $P$ at distance $r$ from $O$[cite: 13].

2. Consider an intermediate point $A$ at distance $x$ from $O$. Electrostatic repulsive force on a test charge $+q_0$ placed at $A$:

$$\vec{F}_e = \frac{1}{4\pi\varepsilon_0}\frac{q q_0}{x^2}\hat{i}$$

3. To move $q_0$ inward by displacement $d\vec{x} = dx\,\hat{i}$ without acceleration, an external agent must apply $\vec{F}_{\text{ext}} = -\vec{F}_e$:

$$dW_{\text{ext}} = \vec{F}_{\text{ext}} \cdot d\vec{x} = \left(-\frac{1}{4\pi\varepsilon_0}\frac{q q_0}{x^2}\right)dx$$

4. Total external work done in bringing $q_0$ from $\infty$ to distance $r$[cite: 13]:

$$W_{\infty \to r}^{\text{ext}} = -\frac{q q_0}{4\pi\varepsilon_0} \int_{\infty}^{r} \frac{1}{x^2}\,dx = -\frac{q q_0}{4\pi\varepsilon_0} \left[ -\frac{1}{x} \right]_{\infty}^{r} = \frac{1}{4\pi\varepsilon_0}\frac{q q_0}{r}$$

5. By definition, electric potential $V = \frac{W^{\text{ext}}}{q_0}$[cite: 13]:

$$V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$$

Derivation B

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. Distances from individual charges for a short dipole ($r \gg a$):

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

3. By algebraic superposition principle of potentials ($V = V_+ + V_-$)[cite: 13]:

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

4. Since $p = q(2a)$, we arrive at the universal relation[cite: 13]:

$$V(r, \theta) = \frac{1}{4\pi\varepsilon_0}\frac{\vec{p}\cdot\hat{r}}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$$

Punchlines: (i) Axial line ($\theta = 0^\circ$): $V_{\text{axial}} = \pm \frac{kp}{r^2}$[cite: 13]. (ii) Equatorial line ($\theta = 90^\circ$): $\cos 90^\circ = 0 \implies V_{\text{eq}} = 0$[cite: 13].

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: "V = 0 implies E = 0" Fallacy

At any point on the equatorial plane of an electric dipole, $V = 0$ identically, yet $\vec{E} \ne 0$ ($E = \frac{kp}{r^3}$ parallel to $-\vec{p}$)[cite: 13]! Similarly, inside a charged spherical shell, $\vec{E} = 0$, yet $V = \frac{kq}{R} \ne 0$[cite: 13]. Always remember: field depends on the spatial slope ($E = -dV/dr$), not on the instantaneous value of $V$ alone.

Trap 2: Forgetting Algebraic Signs in Potential

While calculating $\vec{E}$ or $\vec{F}$, charges are substituted as absolute positive magnitudes because directions are taken care of by unit vectors. But Electric Potential is a Scalar[cite: 13]! You MUST substitute positive charges as $+q$ and negative charges as $-q$ directly into $V = \frac{kq}{r}$[cite: 13].

Trap 3: Work Done by Field vs Work Done by External Agent

$$W_{\text{ext}} = q_0(V_B - V_A) = \Delta U$$ $$W_{\text{electric field}} = -q_0(V_B - V_A) = -\Delta U$$ Examiners frequently swap these terms in objective questions. Always look closely at whether the work done is asked for the electric field or an external agent.

Trap 4: Conservative Path Independence

The work done moving a charge between two points in an electrostatic field is strictly path-independent ($\oint \vec{E}\cdot d\vec{r} = 0$). Moving along a zigzag path of length 10 meters requires the exact same work as a straight line of 2 cm.

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: Two point charges $+3\text{ }\mu\text{C}$ and $-2\text{ }\mu\text{C}$ are located $15\text{ cm}$ apart. At what points on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero[cite: 13].

Solution: Let the line joining them be the $x$-axis with $q_1 = +3\text{ }\mu\text{C}$ at $x = 0$ and $q_2 = -2\text{ }\mu\text{C}$ at $x = 15\text{ cm}$.

Case 1: Between the charges ($0 < x < 15\text{ cm}$):

$$V(x) = \frac{1}{4\pi\varepsilon_0}\left[\frac{3 \times 10^{-6}}{x} + \frac{-2 \times 10^{-6}}{15 - x}\right] = 0 \implies \frac{3}{x} = \frac{2}{15 - x}$$

$$3(15 - x) = 2x \implies 45 = 5x \implies x = 9\text{ cm from }+3\text{ }\mu\text{C}$$

Case 2: Outside the charges (closer to the smaller magnitude charge, $x > 15\text{ cm}$):

$$\frac{3}{x} = \frac{2}{x - 15} \implies 3(x - 15) = 2x \implies 3x - 45 = 2x \implies x = 45\text{ cm from }+3\text{ }\mu\text{C}$$

Verdict: Potential is zero at two distinct points: at $9\text{ cm}$ (between them) and at $45\text{ cm}$ (outside, beyond $-2\text{ }\mu\text{C}$).

JEE/NEET Competitive Trap

Problem: An electric field is given by $\vec{E} = 20x\,\hat{i} + 30y^2\,\hat{j}\text{ (V/m)}$. Find the potential difference $V_O - V_A$, where $O$ is origin $(0,0)$ and point $A$ is $(2\text{ m}, 1\text{ m})$.

Solution: Using line integral $V_A - V_O = -\int_{(0,0)}^{(2,1)} \vec{E}\cdot d\vec{r}$:

Since $d\vec{r} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}$, we have $\vec{E}\cdot d\vec{r} = 20x\,dx + 30y^2\,dy$:

$$V_A - V_O = -\left[ \int_{0}^{2} 20x\,dx + \int_{0}^{1} 30y^2\,dy \right]$$

Evaluate individual integrals:

$$\int_{0}^{2} 20x\,dx = \left[ 10x^2 \right]_0^2 = 10(4) = 40\text{ V}$$

$$\int_{0}^{1} 30y^2\,dy = \left[ 10y^3 \right]_0^1 = 10(1) = 10\text{ V}$$

Thus, $V_A - V_O = -(40 + 10) = -50\text{ V}$.

$$V_O - V_A = -(V_A - V_O) = +50\text{ Volts}$$

🗂️ 07. Quick Revision Flashcard Core Formulae

Point Charge

$V = \frac{kq}{r}$

Dipole Potential

$V = \frac{kp\cos\theta}{r^2}$

Equatorial Dipole

$V_{\text{eq}} = 0$

Field Relation

$E = -\frac{dV}{dr}$