Physics II Study Notes

Physics II Overview

  • Course Structure

    • Credit Hours: 3 Cr. Hrs.

    • Breakdown: 2 hours Lecture (LCT) + 2 hours Laboratory (LAB)

  • Topics Covered:

    • Classical Mechanics

    • Electric Fields

    • Gauss's Law

    • Electric Potential

    • Capacitance and Dielectrics

    • Current and Resistance

    • Direct Current Circuits

    • Magnetic Fields

    • Sources of Magnetic Field

    • Faraday's Law

    • Inductance

    • Alternating Current Circuits

    • Electromagnetic Waves

    • Nature of Light and Laws of Geometric Optics

    • Image Formation

Electric Fields

Definition

  • Electric Field (E):

    • Defined as the space around a charged particle within which a force would be exerted on other charged particles.

Electric Charge

Properties of Electric Charges

  • Types of Electric Charges:

    • Positive (+ve)

    • Negative (-ve)

  • Interaction of Charges:

    • Like charges repel one another

    • Unlike charges attract one another

  • Examples:

    • A negatively charged rubber rod and a positively charged glass rod experience attraction.

    • Two positively charged glass rods will repel one another.

    • Spotlight Concept: Like charges repel each other and unlike charges attract each other.

Conservation of Electric Charges

  • Key Principle:

    • Charge is transferred between objects when one object is rubbed against another; charge is not created in the process.

  • Example:

    • When a glass rod is rubbed with silk, electrons are transferred from the glass to the silk, resulting in negative charge on silk and equal positive charge left on the glass.

  • Spotlight Concept: The total charge in any isolated system is conserved.

Quantization of Electric Charges

  • Definition:

    • Electric charge (q) is quantized, existing in discrete packets.

    • Standard Symbol for Charge:

    • The symbol for charge as a variable is q.

    • It can be expressed as: q=ext±Neq = ext{±} Ne

      • Here, N is an integer and e is the fundamental unit of charge.

      • Fundamental Unit of Charge:

      • ∣e∣=1.6imes10−19C|e| = 1.6 imes 10^{-19} C

      • Electron: q=−eq = -e

      • Proton: q=+eq = +e

Coulomb's Law

Concept

  • Coulomb's Law:

    • Used to measure the magnitude of the electrical force (F) between two charged particles.

    • The equation and description:

    1. The force is inversely proportional to the square of the distance (r) between the charges, directed along the line joining them.

    2. The force is proportional to the product of the charges, q<em>1q<em>1 and q</em>2q</em>2, on the two particles.

  • Mathematical Representation:

    • Feext∝rac1r2F_e ext{∝} rac{1}{r^2}

    • F<em>eext∝q</em>1q2F<em>e ext{∝} q</em>1 q_2

Units and Constants

  • SI Unit of Charge:

    • The unit is the coulomb (C).

  • Coulomb Constant (k):

    • k=8.9876imes109extN⋅m2/extC2k = 8.9876 imes 10^9 ext{ N·m}^2/ ext{C}^2 (rounded to 9).

    • Permittivity of Free Space (ε₀):

    • ε0=8.8542imes10−12extC2/extN⋅m2ε₀ = 8.8542 imes 10^{-12} ext{ C}^2/ ext{N·m}^2

  • Coulomb's Law Formula:

    • F<em>e=kracq</em>1q2r2F<em>e = k rac{q</em>1 q_2}{r^2}

Vector Nature of Electric Forces

  • Vector Quantity:

    • The force is a vector quantity, written as the electric force exerted by charge q<em>1q<em>1 on charge q</em>2q</em>2:

    • extbf{F}{12} = k rac{q1 q2}{r^2} extbf{r{12}}

    • Here, extbfr<em>12extbf{r<em>{12}} is a unit vector directed from q</em>1q</em>1 toward q2q_2.

  • Reciprocal Force Relationship:

    • The electric force exerted by q<em>2q<em>2 on q</em>1q</em>1 is equal in magnitude to the force exerted by q<em>1q<em>1 on q</em>2q</em>2 but in the opposite direction:

    • extbfF<em>21=−extbfF</em>12extbf{F}<em>{21} = - extbf{F}</em>{12}

Multiple Charges

  • Resultant Force on a Charge:

    • The resultant force on a charge equals the vector sum of the forces exerted by all other charges present.

  • Force Components:

    • The components of the resultant force can be expressed in terms of their X and Y parameters:

    • Fx=FextcoshetaF_x = F ext{cos} heta

    • Fy=FextsinhetaF_y = F ext{sin} heta

  • Direction of Force:

    • The overall force direction is determined by the vector sum of the individual forces acting on the charge.

Example Calculations

Example (1)

  • Problem Statement:

    • Three charges are aligned along the X axis:

    • Positive charge q1=15extμCq_1 = 15 ext{ μC} at x=2extmx = 2 ext{ m}

    • Positive charge q2=6extμCq_2 = 6 ext{ μC} at the origin

    • Find the position for a negative charge q3q_3 on the X axis such that the resultant force on it is zero.

Solution Steps
  1. The forces acting on q3q_3 are:

    • F<em>13F<em>{13} and F</em>23F</em>{23} (attractive forces)

  2. Let xx be the coordinate position of q<em>3q<em>3 between q</em>1q</em>1 and q2q_2.

  3. Use the equations:

    • F<em>13=Kracq</em>3q1(2−x)2F<em>{13} = K rac{q</em>3 q_1}{(2-x)^2}

    • F<em>23=Kracq</em>3q2x2F<em>{23} = K rac{q</em>3 q_2}{x^2}

  4. Set the forces equal under the condition F3=0F_3 = 0:

    • F<em>13=F</em>23F<em>{13} = F</em>{23}

  5. Solve the equation:

    • 5(2−x)2=2x25(2 - x)^2 = 2x^2

    • Expanding leads to:

    • 5x2+8x−8=05x^2 + 8x - 8 = 0

  6. Use the quadratic formula to find xx:

    • x=rac−bext±ext√(b2−4ac)2ax = rac{-b ext{±} ext{√}(b^2-4ac)}{2a}

    • Results give: x=0.775extmx = 0.775 ext{ m} and x=−3.442extmx = -3.442 ext{ m}

Example (2)

  • Problem Statement:

    • Two identical small charged spheres each with mass 3imes10−2extKg3 imes 10^{-2} ext{ Kg} hang in equilibrium, with the length of each string 0.15extm0.15 ext{ m} and angle heta=5°heta = 5°. Find the charge on each sphere, assuming they have identical charges.

Solution Steps
  1. Given Data:

    • m=3extx10−2extKgm = 3 ext{ x } 10^{-2} ext{ Kg}

    • L=0.15extmL = 0.15 ext{ m}

    • heta=5°heta = 5°

  2. Find the horizontal distance from the center:

    • a=Lextsinheta=0.15extxextsin(5exto)=0.013extma = L ext{ sin} heta = 0.15 ext{ x } ext{sin}(5^{ ext{o}}) = 0.013 ext{ m}

    • r=2a=0.026extmr = 2a = 0.026 ext{ m}

  3. Forces Acting on One Sphere:

    1. Tension in wire (T)

    2. Weight of sphere (mg)

    3. Coulomb's force (FeF_e)

  4. Since the sphere is in equilibrium, set resultant forces in X and Y directions to zero:

    • extForceinX−axis:Textsin(heta)=Feext{{Force in X-axis}}: T ext{sin}( heta) = F_e

    • extForceinY−axis:Textcos(heta)=mgext{{Force in Y-axis}}: T ext{cos}( heta) = mg

  5. Plugging the equations into Coulomb's Law:

    • Fe=Kracq2r2F_e = K rac{q^2}{r^2}

    • Using the formulas, solve for charge qq and achieve:

    • q=4.4imes10−8extCq = 4.4 imes 10^{-8} ext{ C}

Conclusion

  • Thank you for your attention!