Electrostatics: Charges, Fields, Potential, Capacitors & Applications

Core Concepts & Definitions

1. Electric Charge: The Fundamental Property

Electric charge is a physical property of matter that causes it to experience a force when placed in an electric or magnetic field. Charges exist in two types:

• Positive charge (protons, lost electrons)
• Negative charge (electrons, excess electrons)

classDiagram
    class CoulombsLaw {
      +F = k_e * (|q1 * q2| / r^2)
      +k_e = 9×10^9 N·m²/C²
    }
    class ChargeConservation {
      +Total charge in isolated system = constant
    }
    class Quantization {
      +q = n * e (e = 1.6×10⁻¹⁹ C)
    }
    CoulombsLaw --> ChargeConservation : Governs interactions
    Quantization --> CoulombsLaw : Applies to discrete charges

Key Properties:

• Charge is quantized: $q=n\cdot e$, where $e=1.6\times 10^{-19}\text{C}$ (electron charge).
• Conservation of charge: Total charge in an isolated system remains constant.
• Coulomb’s Law: Force between two point charges is proportional to the product of charges and inversely proportional to the square of the distance between them.

Worked Example (Real-World Tie): In eSewa or Khalti, when you transfer money via a mobile app, the transaction involves electrostatic principles in the underlying hardware (capacitors in circuits) and charge separation in touchscreens. Suppose two users, A and B, have charges $+q$ and $-q$ respectively, separated by 10 cm. Calculate the electrostatic force between them using Coulomb’s law.

Solution: $$F=k_e\frac{|q_A\cdot q_B|}{r^2}=9\times 10^9\cdot \frac{|(+q)(-q)|}{(0.1{)}^2}=9\times 10^9\cdot \frac{q^2}{0.01}=9\times 10^{11}{q}^2\text{N}$$ If $q=1\mu \text{C}=10^{-6}\text{C}$, then $F=9\times 10^5\text{N}$.

2. Electric Field: The "Invisible Force" Around Charges

An electric field ($\vec{E}$) is a region around a charged object where a force is exerted on other charges. It is a vector field (has magnitude and direction).

Electric field lines radiating from a positive point charge, showing field strength (density) and direction. (Image: FormulariosBachillerato, CC0, via Wikimedia Commons)

Key Formulas:

• Electric field due to a point charge: $$\vec{E}=k_e\frac{q}{r^2}\hat{r}$$ where $\hat{r}$ is the unit vector pointing away from the charge (for positive $q$) or toward it (for negative $q$).

• Electric field lines:

  • Start on positive charges, end on negative charges (or at infinity).
  • Never intersect (each point has one direction).
  • Density of lines = strength of the field.

Real-World Example:

• Photocopiers (Xerox machines) use electrostatic attraction to transfer toner (charged particles) onto paper. The drum inside carries a negative charge, attracting positively charged toner particles.
• Nepal’s NTC power lines generate electric fields around high-voltage wires, which must be shielded to prevent interference with electronic devices.

3. Electric Potential & Potential Difference

Electric Potential (V):

The work done per unit charge to bring a test charge from infinity to a point in the field. $$V=k_e\frac{q}{r}$$

• Unit: Volts (V) = Joules/Coulomb (J/C).
• Potential difference (ΔV): Work done per unit charge to move between two points. $$\Delta V=V_B-V_A=-{\int }_A^B\vec{E}\cdot d\vec{l}$$

Equipotential Surfaces:

• Surfaces where potential is constant.
• Perpendicular to electric field lines.
• No work is done moving a charge along an equipotential surface.

Worked Example (Real-World Tie): In Ncell’s base stations, antennas create electric fields to transmit signals. If a charge $q=2\mu \text{C}$ is placed 5 m away from a transmitting antenna (modeled as a point charge), calculate:

1. The electric field at that point.
2. The electric potential at that point.

Solution:

1. Electric field: $$E=k_e\frac{q}{r^2}=9\times 10^9\cdot \frac{2\times 10^{-6}}{(5{)}^2}=7.2\times 10^4\text{N/C}$$
2. Electric potential: $$V=k_e\frac{q}{r}=9\times 10^9\cdot \frac{2\times 10^{-6}}{5}=3.6\times 10^4\text{V}$$

4. Capacitors: Storing Electric Charge

A capacitor is a device that stores electrical energy in an electric field. It consists of two conducting plates separated by an insulating material (dielectric).

Key Formulas:

• Capacitance (C): $$C=\frac{Q}{V}$$ where $Q$ = charge on one plate, $V$ = potential difference between plates.

  • Unit: Farad (F) = Coulombs/Volts (C/V).

• Capacitance of a parallel-plate capacitor: $$C=\frac{{ϵ}_{0}A}{d}$$ where ${ϵ}_0$ = permittivity of free space ($8.85\times 10^{-12}\text{F/m}$), $A$ = plate area, $d$ = separation distance.

• Energy stored in a capacitor: $$U=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{Q^2}{C}$$

Shows two metal plates with separation $d$, connected to a battery, with electric field lines between them. (Image: Geek3, CC BY-SA 4.0, via Wikimedia Commons)

Types of Capacitors & Their Uses:

Real-World Example:

• Daraz’s payment processing uses capacitors to stabilize voltage in servers, ensuring smooth transactions.
• Pathao’s GPS systems rely on capacitors in circuits to filter noise and maintain signal integrity.

5. Dielectrics & Their Effect on Capacitance

A dielectric is an insulating material placed between capacitor plates. It:

• Increases capacitance by reducing the electric field between plates.
• Prevents arcing (breakdown of air).
• Stores more energy for the same voltage.

Key Formula: $$C=\kappa C_0$$ where:

• $C_0$ = capacitance without dielectric,
• $\kappa$ = dielectric constant (dimensionless, $\kappa >1$).

Common Dielectrics & Their $\kappa$ Values:

Shows a parallel-plate capacitor with a dielectric slab inserted, reducing the electric field. (Image: Papa November, CC BY-SA 3.0, via Wikimedia Commons)

Worked Example: A parallel-plate capacitor has plates of area $A=0.01{\text{m}}^2$ separated by $d=1\text{mm}$. If a dielectric with $\kappa =5$ is inserted, and a battery maintains $V=100\text{V}$, find:

1. The capacitance without and with the dielectric.
2. The charge stored in each case.

Solution:

1. Without dielectric ($C_0$): $$C_0=\frac{{ϵ}_{0}A}{d}=\frac{(8.85\times 10^{-12})(0.01)}{0.001}=8.85\times 10^{-11}\text{F}$$ With dielectric ($C$): $$C=\kappa C_0=5\times 8.85\times 10^{-11}=4.425\times 10^{-10}\text{F}$$
2. Charge stored ($Q=CV$):
   • Without dielectric: $Q_0=8.85\times 10^{-11}\times 100=8.85\times 10^{-9}\text{C}$.
   • With dielectric: $Q=4.425\times 10^{-10}\times 100=4.425\times 10^{-8}\text{C}$.

6. Applications of Electrostatics in Technology

A. Everyday Devices

1. Photocopiers & Printers

   • Use electrostatic attraction to transfer toner (charged particles) onto paper.
   • Corona wire charges the drum negatively, attracting positively charged toner.

2. Air Purifiers & HEPA Filters

   • Electrostatic precipitators in air purifiers use charged plates to attract dust particles.

3. Static Cling in Clothes

   • When clothes rub against each other, charge separation occurs, causing attraction (static cling).

B. Modern Electronics

1. Touchscreens (Smartphones, ATMs)

   • Capacitive sensing: Fingers disturb the electric field at the screen’s surface, detecting touch.

2. RAM (Random Access Memory) in Computers

   • Uses capacitors to store bits (0 or 1) as charge states.

3. Flash Memory (USB Drives, SSDs)

   • Relies on charge trapping in floating-gate transistors.

C. Medical Applications

1. Electrostatic Sprayers

   • Used in COVID-19 disinfection (e.g., spraying hospitals with charged droplets for even distribution).

2. Defibrillators

   • Deliver a high-voltage pulse to restart the heart by overriding chaotic electrical signals.

## In the Real World

1. eSewa & Khalti (Digital Payments)

   • Capacitors in payment gateways filter voltage spikes to protect sensitive circuits during transactions.
   • Electrostatic discharge (ESD) protection is critical in servers to prevent data corruption.

2. Nepal’s NTC & Ncell (Power & Telecommunications)

   • High-voltage transmission lines create electric fields that must be managed to avoid corona discharge (wasting energy as light/sound).
   • Capacitors in substations smooth out voltage fluctuations for stable power supply.

3. Daraz & Pathao (Logistics & Ride-Hailing)

   • GPS systems in delivery vehicles use capacitors to stabilize power for accurate location tracking.
   • Static electricity can interfere with sensors; grounding is used to prevent malfunctions.

4. Nepal Stock Exchange (NEPSE) Data Centers

   • Uninterruptible Power Supplies (UPS) use capacitors to provide backup power during outages, ensuring trading systems remain online.

## Exam Tip

What to Focus On:

1. Coulomb’s Law & Electric Field Calculations

   • Always draw a diagram showing charges and distances.
   • Remember: Force is a vector—direction matters!

2. Electric Potential vs. Electric Field

   • Potential (V) is a scalar (no direction), while field (E) is a vector.
   • Use $\Delta V=-\int \vec{E}\cdot d\vec{l}$ for non-uniform fields.

3. Capacitors & Dielectrics

   • Parallel-plate formula: $C=\frac{{ϵ}_{0}A}{d}$ (memorize!).
   • Dielectric effect: $C$ increases by $\kappa$, but voltage drops if battery is disconnected.

4. Real-World Problem Solving

   • eSewa/Khalti: Relate to charge storage in capacitors.
   • NTC power lines: Discuss electric field safety and corona discharge.
   • Touchscreens: Explain capacitive sensing.

Common Mistakes to Avoid:

• Ignoring units: Always check if answers are in N/C (E), V (V), or F (C).
• Sign errors: Potential is positive for positive charges, but field direction is away from positive charges.
• Assuming uniform fields: Only use $E=\frac{V}{d}$ for parallel plates or uniform fields.

High-Score Strategies:

• Draw diagrams for every problem (especially electric fields and capacitors).
• Label all variables in equations (examiners reward clarity).
• Relate to real-world examples (e.g., "This is like the capacitor in a smartphone’s touchscreen").
• Show step-by-step calculations—even if the final answer is wrong, partial marks are possible.

## Quick Revision Table