Physics Semester I Notes: Electrostatics and Coulomb's Law

Review of Mathematical Principles for Physics

Dealing with Exponents

  • Multiplication: To multiply two numbers in scientific notation, multiply their coefficients and add their exponents.

    • Example: (2×103)(4×104)=8×107(2 \times 10^3) (4 \times 10^4) = 8 \times 10^7

  • Division: To divide two numbers in scientific notation, divide their coefficients and subtract their exponents.

    • Example: 4×1022×104=2×102\frac{4 \times 10^2}{2 \times 10^4} = 2 \times 10^{-2}

  • Powers of Exponents: If a number and its exponent are raised by another exponent, you must multiply the exponents.

    • Example: (52)3=56(5^2)^3 = 5^6

  • Conversion: In any case involving these operations, the final answer must be converted to scientific notation.

Fundamental Properties of Charge

Charge Quantities and Units

  • Quantity of Charge: Measured in Coulombs (CC).

  • Elementary Charge Values:

    • 1coulomb=charge of 6.25×1018e1\,\text{coulomb} = \text{charge of } 6.25 \times 10^{18}\,e^-

    • 1e=1.602×1019C1\,e^- = 1.602 \times 10^{-19}\,C

    • 1p=1.602×1019C1\,p = 1.602 \times 10^{-19}\,C

  • Charge Polarity: The charge on the electron is negative, while the charge on the proton is positive. However, their magnitudes are identical.

  • Quantization: Charge cannot be smaller than a single electron. The transfer of charge involves the transfer of discrete electrons.

  • Electron Volt (eVeV): A unit of electrical potential energy where 1eV=1.602×1019J1\,eV = 1.602 \times 10^{-19}\,J.

Coulomb's Law

Charles-Augustin de Coulomb (1736-1806)

  • Coulomb's Law (1785): This is the definitive formula for determining electrical force (FF).

  • Mathematical Formula:     F=k×q1×q2r2F = \frac{k \times q_1 \times q_2}{r^2}

    • FF: Electrical force.

    • kk: The Coulomb constant. Formally, k=8.9875×109Nm2/C2k = 8.9875 \times 10^9\,N \cdot m^2/C^2, though 9×1099 \times 10^9 is commonly used for calculations.

    • q1,q2q_1, q_2: The charges on the respective objects.

    • rr: The distance between the objects.

Conductors and Insulators

Mechanisms of Conductivity

  • Basis of Electricity: Electricity is fundamentally based on the movement of electrons.

  • Valence Electrons: How well an object transfers electricity depends on how strong of a hold it has on its outermost (valence) electrons.

  • Universal Flow: Every substance, with the exception of a total vacuum, can flow current.

Classification of Materials

  • Insulators: These materials have a strong hold on their outermost ee^-.

    • Examples: Rubber, plastic, wood, glass, Styrofoam, air.

    • Special Cases: Vacuums contain no electrons, and distilled water contains no loose ions or electrons to facilitate flow.

  • Conductors: These materials have a loose hold on their outermost ee^-.

    • Examples: Metals (copper, gold, aluminum, etc.) and water (specifically salt water).

  • Semiconductors: These materials can switch between being very good conductors and very good insulators. This state change is achieved by altering a very small number of atoms.

    • Example: Silicon, which acts like a switch in computer chips.

  • Superconductivity: Materials that can flow a near-infinite amount of current.

    • Conditions: Occurs only at extremely low temperatures, ranging from near absolute zero to approximately 100K100\,K.

    • Applications: Used in high-performance computers, internet infrastructure, and magnetic levitation.

Methods of Charging

Charging by Friction

  • Process: This involves knocking the outermost valence electrons off from conductors or materials with a loose hold on electrons.

  • Examples:

    • Walking across a carpet in bunny slippers.

    • Rubbing a balloon against one's hair.

Charging by Contact

  • Process: Physical contact is made between two objects, where at least one has an excess of charge.

    • Example: Touching a doorknob after becoming "charged up" from a floor via bunny slippers.

  • Electron Behavior: Electrons do not want to stay together because they share the same charge. They will move toward better conductors or larger objects where they can spread out more effectively.

Charging by Induction and Charge Polarization

  • Induction: This occurs when a charged object is brought near an uncharged object. The electrical field of the charged object induces a charge in the uncharged object without physical contact.

  • Charge Polarization:

    • Interaction: The electrical field of the charged object interacts with the uncharged object.

    • Separation: Electrons move to one side, creating positive and negative poles in the object.

    • Attraction: For example, if a negatively charged object is near, electrons move away, leaving protons behind; the now positively charged side is attracted to the negatively charged object.

  • Shape and Distribution: Charge will build up or spread across an object based on its shape. "Pointed" objects typically build up charge on their ends.

Electrical Fields

Field Strength and Calculation

  • Definition: A build-up of charge produces an electrical field. The strength of this field is determined by the number of excess electrons or protons involved.

  • Formula for Electrical Field Strength:     E=FqE = \frac{F}{q}

    • EE: Electrical field strength, measured in Newtons per Coulomb (N/CN/C).

    • FF: Force, typically measured in Newtons (NN).

    • qq: Quantity of charge, measured in Coulombs (CC).

Atmospheric Phenomena: Lightning

The Formation of Lightning

  • Origins: Lightning is a primary example of both charging by friction and charging by induction.

  • Development: In a thundercloud, massive updrafts occur due to pressure changes (e.g., a cold front undercutting a warm front, causing it to rise rapidly).

  • Friction and Cooling: Rapid rising causes rapid cooling and the development of ice particles. Smaller ice crystals rub against larger ice crystals as they rise, facilitating charging by friction.

  • Charge Distribution within Clouds: Small crystals (rising) lead to a massive shortage of electrons at the top of the cloud. Large crystals (falling) create a massive excess of electrons at the bottom of the cloud.

  • Internal Discharge: When the charge build-up becomes too great, "sheet" lightning occurs within the cloud.

Cloud-to-Ground Strikes

  • Ground Induction: The excess negative charge at the cloud's bottom induces a charge in the ground. Loose electrons in the ground move away, leaving behind a concentration of protons.

  • The Strike: A "negative leader" from the cloud connects with a "positive leader" from the ground, resulting in a lightning strike to the ground.

  • Safety Warning: One should never become the pathway to the ground for this electrical discharge.


To apply Coulomb's Law and the calculation of electrical field strength in real-world problems, consider the following steps using simple example numbers:

Coulomb's Law

Formulation:
F=k×q<em>1×q</em>2r2F = \frac{k \times q<em>1 \times q</em>2}{r^2}
Where:

  • FF is the electrical force (in Newtons, NN).
  • kk is the Coulomb constant, approximately 8.9875×109Nm2/C28.9875 \times 10^9 N \cdot m^2/C^2 (often simplified to 9×1099 \times 10^9).
  • q<em>1q<em>1 and q</em>2q</em>2 are the magnitudes of the charges (in Coulombs, CC).
  • rr is the distance between the charges (in meters, mm).

Example Problem:
Calculate the electric force between two charges:

  • Charge 1: q1=1imes106Cq_1 = 1 imes 10^{-6} C
  • Charge 2: q2=2imes106Cq_2 = 2 imes 10^{-6} C
  • Distance: r=0.1mr = 0.1 m

Step 1: Substitute values into the formula:
F=9×109×(1×106)×(2×106)(0.1)2F = \frac{9 \times 10^9 \times (1 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.1)^2}

Step 2: Calculate:

  • Multiply the charges: (1×2)=2×1012(1 \times 2) = 2 \times 10^{-12}
  • Square the distance: (0.1)2=0.01(0.1)^2 = 0.01
  • Substitute back into formula:
    F=9×109×2×10120.01F = \frac{9 \times 10^9 \times 2 \times 10^{-12}}{0.01}
  • Calculate the force:
    F=18×1030.01=1.8×103NF = \frac{18 \times 10^{-3}}{0.01} = 1.8 \times 10^3 N
Electrical Field Strength

Formulation:
E=FqE = \frac{F}{q}
Where:

  • EE is the electrical field strength (in Newtons per Coulomb, N/CN/C).
  • FF is the force calculated (in Newtons, NN).
  • qq is the charge experiencing the field (in Coulombs, CC).

Example Problem:
Using the force calculated above (F=1.8×103NF = 1.8 \times 10^3 N) and a charge of q=1×106Cq = 1 \times 10^{-6} C:

Step 1: Substitute into the formula:
E=1.8×1031×106E = \frac{1.8 \times 10^3}{1 \times 10^{-6}}

Step 2: Calculate the field strength:
E=1.8×103×106=1.8×109N/CE = 1.8 \times 10^3 \times 10^6 = 1.8 \times 10^9 N/C

Conclusion

By applying these formulas effectively, one can determine the forces between charges and the strength of electric fields in practical scenarios.