JEE Physics Power Batch Lecture 03: Coulomb's Law and Electrostatic Force Properties

Course and Lecture Information

  • Batch: POWER BATCH (JEE)
  • Subject: Physics
  • Topic: Concept Building
  • Lecture Number: Lecture No-03
  • Instructor: Abhishek Lahoti Sir

Recap of Previous Lecture

  • Introduction: Initial overview of electrostatics.
  • Electric Charges: Fundamental concepts and definitions.
  • Basic Properties of Electric Charge: Quantization, conservation, and additivity.
  • Charging Methods: Friction, conduction, and induction.

Current Lecture Topics

  • Coulomb's Law: Definition, mathematical derivation, and limitations.
  • Properties of Coulomb's Law: Essential characteristics and physical implications.

Conceptual Questions and Basic Calculations

  • Proton vs. Electron Charge Comparison [HCV]:     - Question: The charge on a proton is +1.6×1019C+1.6 \times 10^{-19}\,C and that on an electron is 1.6×1019C-1.6 \times 10^{-19}\,C. Does it mean that the electron has a charge 3.2×1019C3.2 \times 10^{-19}\,C less than the charge of a proton?     - Conceptual Insight: The positive and negative signs represent the nature (polarity) of the charge, not algebraic value. Magnitude-wise, they are equal (1.6×1019C1.6 \times 10^{-19}\,C).
  • Charge on an α\alpha-particle [NCERT]:     - An α\alpha-particle is a Helium nucleus expressed as 24He2+{}_{2}^{4}He^{2+}.     - The charge of an α\alpha-particle is equivalent to 22 protons: q=+2eq = +2e.     - Calculation: Charge=2×1.6×1019C=3.2×1019C\text{Charge} = 2 \times 1.6 \times 10^{-19}\,C = 3.2 \times 10^{-19}\,C.     - Options Provided:         - A: 4.8×1019C4.8 \times 10^{-19}\,C         - B: 1.6×1019C1.6 \times 10^{-19}\,C         - C: 3.2×1019C3.2 \times 10^{-19}\,C (Correct Answer)         - D: 6.4×1019C6.4 \times 10^{-19}\,C

Coulomb's Law

  • Definition: The magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
  • Mathematical Representation:     - Consider two point charges q1q_1 and q2q_2 separated by a distance rr.     - Fq1q2F \propto q_1 q_2     - F1r2F \propto \frac{1}{r^2}     - Combining these: F=kq1q2r2F = \frac{k q_1 q_2}{r^2}
  • Scope of Application: This law is strictly applicable only to point charges.
  • Interaction Direction:     - The force between charges acts along the line joining the centers of the charges.     - F12F_{12} is the force on charge q1q_1 due to q2q_2.     - F21F_{21} is the force on charge q2q_2 due to q1q_1.     - Magnitude: F12=F21=F|F_{12}| = |F_{21}| = F.     - Direction: They are always in opposite directions (F12=F21F_{12} = -F_{21}).

Proportionality Constant and Permittivity

  • Constant of Proportionality (kk):     - k=14πϵk = \frac{1}{4 \pi \epsilon}, where ϵ\epsilon is the permittivity of the medium.
  • In Vacuum (Free Space):     - Permittivity of free space ϵ0=8.85×1012C2N1m2\epsilon_0 = 8.85 \times 10^{-12}\,C^2 N^{-1} m^{-2}.     - k=14πϵ0=9×109Nm2C2k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9\,N m^2 C^{-2}.
  • In a Medium:     - Permittivity is defined as ϵ=ϵ0ϵr\epsilon = \epsilon_0 \epsilon_r.     - ϵr\epsilon_r is the Relative Permittivity (also known as the Dielectric Constant).

Relative Permittivity (ϵr\epsilon_r)

  • Ratio Definition: ϵr=ϵϵ0\epsilon_r = \frac{\epsilon}{\epsilon_0}.
  • Dimension: Since it is a ratio of similar quantities, relative permittivity is dimensionless.
  • Values for Common Media:     - Vacuum: ϵr=1\epsilon_r = 1     - Air: ϵr1\epsilon_r \approx 1     - Water: ϵr=81\epsilon_r = 81     - Metals/Conductors: ϵr=\epsilon_r = \infty

Dimensional Analysis

  • Dimension of kk:     - From F=kq1q2r2F = \frac{k q_1 q_2}{r^2}, we get k=Fr2q2k = \frac{F r^2}{q^2}.     - [k]=[MLT2][L2][AT]2=[ML3T4A2][k] = \frac{[M L T^{-2}] [L^2]}{[A T]^2} = [M L^3 T^{-4} A^{-2}].
  • Dimension of ϵ\epsilon:     - Since k=14πϵk = \frac{1}{4 \pi \epsilon}, the dimensions of ϵ\epsilon are the inverse of kk.     - [ϵ]=[M1L3T4A2][\epsilon] = [M^{-1} L^{-3} T^4 A^2].

Electrostatic Force in Mediums

  • Force Reduction Principle: The electrostatic force between two charges decreases when they are placed in a medium other than vacuum.
  • Formula for Force in Medium (FmF_m):     - Fm=14πϵ0ϵrq1q2r2F_m = \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}     - Fm=FvacuumϵrF_m = \frac{F_{vacuum}}{\epsilon_r}
  • Key Observation: Since ϵr1\epsilon_r \ge 1, it follows that FmFvacuumF_m \le F_{vacuum}.
  • Example 1: If force in vacuum Fv=100NF_v = 100\,N and medium relative permittivity ϵr=20\epsilon_r = 20, then force in medium Fm=10020=5NF_m = \frac{100}{20} = 5\,N.
  • Example 2:     - Given: Force in vacuum Fv=48NF_v = 48\,N.     - Question A: Find force in medium with ϵr=3\epsilon_r = 3.     - Calculation: Fm=483=16NF_m = \frac{48}{3} = 16\,N.     - Question B: If force in medium is found to be 12N12\,N, find the relative permittivity ϵr\epsilon_r.     - Calculation: ϵr=FvFm=4812=4\epsilon_r = \frac{F_v}{F_m} = \frac{48}{12} = 4.

Physical Properties of Electrostatic Force

  • Independence Principle: The electrostatic force of attraction or repulsion between two specific charges is independent of the presence or absence of other surrounding charges. (Note: While individual pair forces remain constant, the net force on a charge may change).
  • Central Force Nature:     - The force acting on one point charge due to another is always along the line joining these two charges.     - It is equal in magnitude and opposite in direction regardless of the magnitudes of the charges (Newton's Third Law).     - Electrostatic force is characterized as a central force.
  • Conservative Nature:     - Electrostatic force is a conservative force.     - The Work Done (WDW D) by the electrostatic force in moving a point charge along any closed loop shape is zero.

Relationship Between Force and Potential Energy

  • The electrostatic force (F\mathbf{F}) is related to the Potential Energy (UU) via the negative gradient:     - F=U=[Uxi^+Uyj^+Uzk^]\mathbf{F} = -\nabla U = - \left[ \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} + \frac{\partial U}{\partial z} \hat{k} \right]
  • Mathematical Example:     - Let U=x2y3zU = x^2 y^3 z.     - Partial derivative with respect to xx: Ux=2xy3z\frac{\partial U}{\partial x} = 2 x y^3 z     - Partial derivative with respect to yy: Uy=3x2y2z\frac{\partial U}{\partial y} = 3 x^2 y^2 z     - Partial derivative with respect to zz: Uz=x2y3\frac{\partial U}{\partial z} = x^2 y^3     - Thus, F=(2xy3zi^+3x2y2zj^+x2y3k^)\mathbf{F} = -(2 x y^3 z \hat{i} + 3 x^2 y^2 z \hat{j} + x^2 y^3 \hat{k}).