Fundamentals of Electric Charge (Quantization, Conservation, Additivity)
Electric Charge is the intrinsic property of matter that gives rise to electric and magnetic forces and interactions. In the SI system, charge is measured in Coulombs (C).
Basic properties of electric charge:
- Quantization of Charge: Charge on any object is an integral multiple of the elementary charge e = 1.602 × 10-19 C:
q = ± n e (where n = 1, 2, 3, … and e = 1.602 × 10-19 C)
- Conservation of Charge: The algebraic sum of positive and negative charges in an isolated system remains constant over time.
- Additivity of Charge: Total charge of a system is the simple scalar sum of all individual charges: qtotal = q1 + q2 + … + qn.
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Coulomb's Law in Scalar and Vector Form with Dielectric Effects
Coulomb's Law states that the electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:
Where:
- ε₀ = 8.854187 × 10-12 C2/(N·m2) (Permittivity of Free Space).
- k = 1 / (4πε₀) ≈ 8.988 × 109 N·m2/C2.
- Dielectric Medium Effect: When placed in a medium of dielectric constant K (relative permittivity εr), the force is reduced:
Fmed = F₀ / K = (1 / (4πε₀ K)) × (|q1 q2| / r2)
Coulomb's Law in Vector Form:
This confirms that electrostatic forces obey Newton's Third Law of Motion (action-reaction pairs).
Electric Field Intensity, Superposition Principle, and Field Line Properties
The Electric Field (E) at a point is defined as the electrostatic force experienced per unit positive test charge placed at that point:
Key Properties of Electric Field Lines:
- Field lines originate from positive charges and terminate on negative charges.
- Tangent at any point on a field line gives the direction of the electric field at that point.
- Two field lines never intersect because if they do, there would be two different directions of electric field at the point of intersection, which is physically impossible.
- Electrostatic field lines do not form closed loops because electrostatic forces are conservative.
Electric Dipole Derivations (Axial Field, Equatorial Field, Torque and Potential Energy)
An Electric Dipole consists of a pair of equal and opposite point charges (+q and -q) separated by a small distance 2a. The Electric Dipole Moment (p) is a vector directed from -q to +q:
| Dipole Configuration | Exact Mathematical Formula | Short Dipole Approximation (r ≫ a) |
|---|---|---|
| Axial Point (End-on) | Eaxial = (1) / (4piε₀) (2pr) / ((r2 - a2)2) | Eaxial = (2kp) / (r3) (Along p) |
| Equatorial Point (Broadside-on) | Eeq = (1) / (4piε₀) (p) / ((r2 + a2)3/2) | Eeq = (kp) / (r3) (Opposite to p) |
| Ratio (Eaxial / Eeq) | — | Eaxial = 2 Eeq |
| Torque in Uniform Field | τ = p × E | τ = pE sinθ (Max at θ = 90^circ) |
| Potential Energy | U = -p·E | U = -pE cosθ (Stable equilibrium at θ = 0^circ) |
Gauss's Law and Master Derivations (Line Wire, Plane Sheet, Spherical Shell)
Gauss's Law states that the total electric flux (ΦE) passing through any closed Gaussian surface equals the net charge enclosed divided by ε₀:
Electric Field Due to Infinitely Long Straight Charged Wire
Using a cylindrical Gaussian surface of radius r and length l enclosing charge q = λ l:
Electric Field Due to Uniformly Charged Infinite Plane Sheet
Using a cylindrical Gaussian pillbox of cross-sectional area A:
Electric Field Due to Uniformly Charged Thin Spherical Shell
- Outside the shell (r > R): E = (1) / (4πε₀) (q) / (r2) (Behaves as if all charge is at the center).
- On the surface (r = R): E = (1) / (4πε₀) (q) / (R2) = σ / ε₀.
- Inside the shell (r < R): E = 0 (Since qenclosed = 0).
Step-by-Step Solved Board Numerical Problems
Problem 1: Coulomb Force with Dielectric Medium
Two point charges q1 = +2 μC and q2 = -3 μC are placed 10 cm apart in a medium with dielectric constant K = 5. Calculate the electrostatic force.
F = (1 / (4πε₀ K)) × (|q1 q2| / r2) = ((9 × 109) / 5) × ((2 × 10-6)(3 × 10-6) / (0.1)2)
F = 1.8 × 109 × (6 × 10-12 / 0.01) = 1.08 N (Attractive).
Problem 2: Gauss's Law Flux through a Cube
A point charge of 8.854 nC is placed at the center of a cube of edge 10 cm. Find the total flux and flux through one face.
Flux through one face Φface = Φtotal / 6 = 1000 / 6 = 166.67 N·m2/C.
High-Yield CBSE Board Examination Tips and Common Pitfalls
- Vector Signs: In Coulomb's law magnitude calculations, do not substitute negative signs inside |q1 q2|; state direction explicitly as attractive or repulsive.
- Zero Field Inside Shell: Remember that E = 0 inside a conducting shell, but electric potential is constant (V = (kq) / (R)).
- Dipole Direction: Dipole moment p is directed from -q to +q, while electric field is from +q to -q.
Frequently Asked Questions (FAQs)
Quantization of charge states that the total charge of any body is always an integral multiple of a basic quantum unit of charge: q = ± n e, where e = 1.602 × 10-19 C and n is an integer (1, 2, 3…).
In vector notation, the electrostatic force exerted by charge q1 on q2 is: F21 = (1) / (4piε₀) (q1 q2) / (r2) r12, where r12 is the unit vector directed from q1 to q2.
For a short electric dipole of dipole moment p at a distance r: Eaxial = (2kp) / (r3) and Eequatorial = (kp) / (r3). Thus, Eaxial = 2 × Eequatorial at the same distance r.
Gauss's Law states that the total electric flux through any closed Gaussian surface in vacuum is equal to 1/ε₀ times the net charge enclosed inside the surface: ∮ E · dA = frac{qenclosed}{ε₀}.
The electric field at any point inside a uniformly charged spherical shell is identically zero (E = 0) because a Gaussian surface inside the shell encloses zero net charge (qenclosed = 0).
When a dielectric material of relative permittivity (dielectric constant) K is introduced between charges, the electrostatic force decreases by a factor of K: Fmedium = frac{Fvacuum}{K}.