Chapter 01 · Comprehensive Board Reference
Electric Charges & Fields: The Complete Masterclass.
Exhaustive NCERT derivations, vector Coulomb interactions, dipole torque calculations, 5-mark solved board exemplars, and 20 practice MCQs. Built for rigorous Class 12 board examination preparation.
"Every equals sign in a derivation must earn its step marks." — Jatin Sandilya
Engineered by Jatin Sandilya · Step-by-Step Pedagogical Flow
every charge alters the universe
vector direction is
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100% NCERT ScopeDerivations in Full Sequence
5 Complete ProofsGauss & Dipole Theorems
5-Mark ExemplarsSolved with Board Marking Keys
20 Live MCQsAssertion & Numerical Types
every equation runs in reality
How Electrostatics Operates in Daily Life.
How Electrostatics Operates in Daily Life.
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Application 01 · Industry
Photocopiers & Laser Printers
How does toner stick only to letters on a page? Electrostatic attraction dictates every printed character.
Coulomb Force
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The Electrostatic Drum
"Opposite charges attract with precise microscopic forces."
A laser beam discharges specific areas of a selenium drum. Oppositely charged black toner powder clings only to the charged image pattern before being melted onto paper.
Application 02 · Safety
Lightning Protection & Aircraft
Why are passengers inside a metal airplane or car completely safe during a severe lightning strike?
Gauss's Theorem
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Electrostatic Shielding
"Inside a closed conducting shell, the electric field is strictly zero."
Lightning strikes the exterior metal skin of the plane. All charge stays on the outer surface and flows harmlessly away. Zero flux penetrates the cabin.
1. Fundamentals of Electric Charge
Electric Charge is an intrinsic scalar physical property of elementary matter (electrons and protons) causing it to experience an electrostatic force in an electromagnetic field. Its SI unit is Coulomb ($C$), with dimensional formula $[M^0 L^0 T^1 A^1]$.
- Polarity of Charge: The property distinguishing positive (electron deficit) and negative (electron excess) charges, discovered by Benjamin Franklin.
- Gold-Leaf Electroscope: Apparatus detecting presence and nature of electric charge. Leaf divergence angle reflects charge magnitude.
- Methods of Charging:
- Friction: Work-function mismatch causes electron transfer when rubbing two insulators (e.g. glass rod on silk).
- Conduction: Direct physical contact transfer between conductors until potentials equalize.
- Induction: Redistribution of internal charges caused by an external field without touching, followed by grounding.
2. Conductors, Insulators & Semiconductors
| Property | Conductors | Insulators (Dielectrics) | Semiconductors |
|---|---|---|---|
| Free Carrier Density | High free electron density ($\sim 10^{28}/\text{m}^3$) | Negligible free electrons; bound valence electrons | Moderate; increases exponentially with temperature |
| Charge Distribution | Resides solely on the outer boundary surface | Remains localized at point of application | Regulated by intentional doping ($p$, $n$) |
| Examples | Copper, Aluminium, Human Body | Glass, Rubber, Mica, Dry Wood | Silicon, Germanium, GaAs |
3. Three Core Postulates of Electric Charge
- Additivity: Total charge is the scalar algebraic sum: $Q_{\text{net}} = \sum q_i$.
- Conservation: Net charge in an electrically isolated system remains strictly constant over time ($Q_{\text{initial}} = Q_{\text{final}}$).
- Quantisation: Charge exists exclusively in integer multiples of elementary quantum charge $e = 1.602 \times 10^{-19}\text{ C}$:
Quantisation Formula$$q = \pm n e \quad \text{where } n \in \mathbb{N} = \{1, 2, 3, \dots\}$$At macroscopic scales ($q \sim 1\,\mu\text{C} \approx 6.25 \times 10^{12}e$), discrete steps blur into a continuous fluid.
4. Coulomb's Law (Scalar & Vector Proofs)
Coulomb's Law states that the electrostatic force between two stationary point charges is directly proportional to their product and inversely proportional to the square of their separation distance $r$.
Scalar Magnitude
$$F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}$$
Permittivity of vacuum $\varepsilon_0 = 8.854 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$; $\frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$.
Board Derivation 1: Coulomb's Law in Vector Form
Step 1 Let displacement vector pointing from $q_1$ to $q_2$ be $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1$, with unit vector $\hat{r}_{21} = \frac{\vec{r}_{21}}{r}$.
Step 2 The electrostatic force on $q_2$ exerted by $q_1$ is:
$$\vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1)$$
Step 3 Force on $q_1$ exerted by $q_2$ with $\hat{r}_{12} = -\hat{r}_{21}$:
$$\vec{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{12} = -\vec{F}_{21}$$
Conclusion: Electrostatic force satisfies Newton's Third Law and acts along the central axis (central conservative force).
5. Dielectric Constant ($K$ or $\varepsilon_r$)
In a dielectric medium, Coulomb force is reduced by medium polarization:
$$F_{\text{med}} = \frac{F_{\text{vac}}}{K} \implies K = \varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$$Vacuum $K=1$; air $K \approx 1.00054$; water $K \approx 80$; ideal metals $K = \infty$.
1. Electric Field Intensity ($\vec{E}$)
The electrostatic force experienced per unit infinitesimal positive test charge:
Field Equation
$$\vec{E}(\vec{r}) = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{r}$$
SI Unit: $\text{N/C}$ or $\text{V/m}$. Dimensions: $[M^1 L^1 T^{-3} A^{-1}]$. Radially outward for $+Q$; inward for $-Q$.
2. Electric Dipole & Dipole Moment Vector ($\vec{p}$)
Pair of equal and opposite charges ($\pm q$) separated by distance $2\vec{a}$:
Dipole Moment
$$\vec{p} = q \cdot (2\vec{a}) \quad (\text{Directed from } -q \text{ to } +q)$$
Board Derivation 2: Field of a Dipole at an Axial Point ($P$)
Step 1 Place dipole along $x$-axis centered at $O$. Observation point $P$ is at distance $r$ along the axis.
Step 2 Distance from $+q$ is $(r - a)$ and from $-q$ is $(r + a)$:
$$E_{\text{axial}} = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{4ar}{(r^2 - a^2)^2} \right]$$
Step 3 For short dipole ($r \gg a$):
$$\vec{E}_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2\vec{p}}{r^3} \quad (\text{Directed along } \vec{p})$$
Board Derivation 3: Field of a Dipole at an Equatorial Point ($Q$)
Step 1 Let point $Q$ be on the perpendicular bisector at distance $r$. Distance from each charge is $\sqrt{r^2 + a^2}$.
Step 2 Vertical sine components cancel. Horizontal cosine components reinforce in direction opposite to $\vec{p}$:
$$E_{\text{eq}} = 2 E_{+q} \cos\theta = 2 \left( \frac{q}{4\pi\varepsilon_0 (r^2 + a^2)} \right) \left( \frac{a}{\sqrt{r^2 + a^2}} \right) = \frac{q \cdot 2a}{4\pi\varepsilon_0 (r^2 + a^2)^{3/2}}$$
Step 3 For short dipole ($r \gg a$):
$$\vec{E}_{\text{equatorial}} = -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}}{r^3} \quad (\text{Directed opposite to } \vec{p})$$
Key Golden Ratio: At equal distances $r$, $|\vec{E}_{\text{axial}}| = 2 \times |\vec{E}_{\text{equatorial}}|$.
3. Dipole in Uniform Field ($\vec{E}$)
Torque & Potential Energy
$$\vec{\tau} = \vec{p} \times \vec{E} \implies \tau = pE\sin\theta$$
$$U(\theta) = -\vec{p}\cdot\vec{E} = -pE\cos\theta$$
Stable equilibrium at $\theta = 0^\circ$ ($U_{\min} = -pE$). Unstable at $\theta = 180^\circ$ ($U_{\max} = +pE$).
1. Gauss's Law
Gauss Equation
$$\oint_S \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \Phi_{\text{net}}$$
Flux depends strictly on net enclosed charge, not on the geometric shape or size of the Gaussian surface.
Board Derivation 4: Field of an Infinite Uniformly Charged Straight Wire
Step 1 Consider an infinite line of linear charge density $\lambda$. Choose a coaxial cylindrical Gaussian surface of radius $r$ and length $l$.
Step 2 Flux through circular caps is zero ($\vec{E} \perp d\vec{A}$). On curved mantle: $\vec{E} \parallel d\vec{A}$.
$$\oint \vec{E} \cdot d\vec{A} = E (2\pi r l) = \frac{\lambda l}{\varepsilon_0} \implies E = \frac{\lambda}{2\pi\varepsilon_0 r}$$
Board Derivation 5: Field of an Infinite Thin Plane Sheet
Step 1 Let surface charge density be $\sigma$. Choose a Gaussian cylinder of cross-sectional area $A$ piercing perpendicularly through the sheet.
Step 2 Flux emerges through both end-caps: $2EA = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}$ (Independent of distance $r$).
Board Derivation 6: Field of a Thin Spherical Shell (Radius $R$)
Case A Outside ($r > R$): $E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$.
Case B Surface ($r = R$): $E = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2} = \frac{\sigma}{\varepsilon_0}$ (Maximum).
Case C Inside ($r < R$): Enclosed charge $q_{\text{enc}} = 0 \implies E_{\text{inside}} = 0$.
5 Marks • Superposition & Equilibrium
CBSE Board Exemplar
Two point charges $+4e$ and $+e$ are fixed at distance $a$ apart in vacuum. Where should a third charge $q$ be placed on the line joining them so that the whole system remains in equilibrium? Find the magnitude and sign of $q$.
Step 1: Equilibrium of third charge $q$:
Let $q$ be placed at distance $x$ from $+4e$. Distance from $+e$ is $(a - x)$.
$$\frac{1}{4\pi\varepsilon_0} \frac{(4e)q}{x^2} = \frac{1}{4\pi\varepsilon_0} \frac{(e)q}{(a - x)^2} \implies \frac{2}{x} = \frac{1}{a - x} \implies x = \frac{2a}{3}$$
Step 2: Equilibrium of $+e$ for system stability:
$$\frac{1}{4\pi\varepsilon_0} \frac{(4e)(e)}{a^2} + \frac{1}{4\pi\varepsilon_0} \frac{q(e)}{(a/3)^2} = 0 \implies \frac{4e^2}{a^2} + \frac{9qe}{a^2} = 0 \implies q = -\frac{4e}{9}$$
Final Result: Place $q = -\frac{4e}{9}$ at distance $\frac{2a}{3}$ from $+4e$.
Let $q$ be placed at distance $x$ from $+4e$. Distance from $+e$ is $(a - x)$.
$$\frac{1}{4\pi\varepsilon_0} \frac{(4e)q}{x^2} = \frac{1}{4\pi\varepsilon_0} \frac{(e)q}{(a - x)^2} \implies \frac{2}{x} = \frac{1}{a - x} \implies x = \frac{2a}{3}$$
Step 2: Equilibrium of $+e$ for system stability:
$$\frac{1}{4\pi\varepsilon_0} \frac{(4e)(e)}{a^2} + \frac{1}{4\pi\varepsilon_0} \frac{q(e)}{(a/3)^2} = 0 \implies \frac{4e^2}{a^2} + \frac{9qe}{a^2} = 0 \implies q = -\frac{4e}{9}$$
Final Result: Place $q = -\frac{4e}{9}$ at distance $\frac{2a}{3}$ from $+4e$.
5 Marks • Dipole & Gauss Flux
CBSE Core Question
An electric dipole has charges $\pm 2\,\mu\text{C}$ separated by $4\,\text{cm}$. Calculate:
(i) Dipole moment $\vec{p}$.
(ii) Field at $20\,\text{cm}$ on the axial line.
(iii) Field at $20\,\text{cm}$ on the equatorial line.
(iv) Total flux through a concentric sphere of radius $10\,\text{cm}$.
(i) Dipole moment $\vec{p}$.
(ii) Field at $20\,\text{cm}$ on the axial line.
(iii) Field at $20\,\text{cm}$ on the equatorial line.
(iv) Total flux through a concentric sphere of radius $10\,\text{cm}$.
(i) Dipole Moment: $p = q(2a) = (2 \times 10^{-6}\text{ C})(0.04\text{ m}) = \mathbf{8 \times 10^{-8}\text{ C}\cdot\text{m}}$ ($-q$ to $+q$).
(ii) Axial Field ($r = 0.2\text{ m}$):
$$E_{\text{axial}} = \frac{2kp}{r^3} = \frac{2 \times (9 \times 10^9) \times (8 \times 10^{-8})}{(0.2)^3} = \mathbf{1.8 \times 10^5\text{ N/C}}$$
(iii) Equatorial Field: $E_{\text{eq}} = \frac{E_{\text{axial}}}{2} = \mathbf{9.0 \times 10^4\text{ N/C}}$.
(iv) Total Flux: Enclosed charge $q_{\text{enc}} = (+2\,\mu\text{C}) + (-2\,\mu\text{C}) = 0 \implies \Phi_E = \mathbf{0\text{ N}\cdot\text{m}^2/\text{C}}$.
(ii) Axial Field ($r = 0.2\text{ m}$):
$$E_{\text{axial}} = \frac{2kp}{r^3} = \frac{2 \times (9 \times 10^9) \times (8 \times 10^{-8})}{(0.2)^3} = \mathbf{1.8 \times 10^5\text{ N/C}}$$
(iii) Equatorial Field: $E_{\text{eq}} = \frac{E_{\text{axial}}}{2} = \mathbf{9.0 \times 10^4\text{ N/C}}$.
(iv) Total Flux: Enclosed charge $q_{\text{enc}} = (+2\,\mu\text{C}) + (-2\,\mu\text{C}) = 0 \implies \Phi_E = \mathbf{0\text{ N}\cdot\text{m}^2/\text{C}}$.
Q1. If $10^{19}$ electrons are removed from a neutral copper sphere, the net charge acquired is:
Q2. Two charges repel with force $F$ in vacuum. If distance is doubled and medium replaced with water ($K=81$), new force is:
Q3. What is the SI unit and dimensional formula of permittivity of free space ($\varepsilon_0$)?
Q4. A soap bubble is given negative electric charge. Its radius will immediately:
Q5. An electric dipole is inside a closed cube. Total electric flux through all six faces is:
Q6. Ratio of axial field to equatorial field of a short dipole at equal distance $r$ is:
Q7. An electric dipole $\vec{p}$ is in stable equilibrium in field $\vec{E}$ at angle:
Q8. How does electric field $E$ due to an infinite uniformly charged wire vary with distance $r$?
Q9. The electric field $E$ due to an infinite plane thin charged sheet is:
Q10. The electric field inside a hollow spherical charged conductor at $r < R$ is:
Q11. Charge $+q$ is at the center of an open hemispherical bowl. Flux through curved surface is:
Q12. If entering flux is $\Phi_1$ and leaving flux is $\Phi_2$, net enclosed charge is:
Q13. Which statement regarding electric field lines is INCORRECT?
Q14. The torque $\vec{\tau}$ experienced by an electric dipole $\vec{p}$ in field $\vec{E}$ is:
Q15. Work done in rotating a dipole from $\theta = 0^\circ$ to $\theta = 90^\circ$ in field $E$ is:
Q16. Two point charges $+4q$ and $+q$ are separated by distance $L$. Third charge $Q$ is in equilibrium at:
Q17. Why cannot an electrostatic field exist tangentially along a conductor in equilibrium?
Q18. Direction of electric dipole moment vector $\vec{p}$ is conventionally:
Q19. How many electrons constitute $1\text{ C}$ of negative charge?
Q20. If radius of a Gaussian sphere enclosing charge $Q$ is doubled, outward flux will:
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