03Electrostatics
Analyze electric charge, apply Coulomb's law, describe electric fields and potential, and distinguish conductors from insulators.
Electrostatics is the foundation of all electrical phenomena. Understanding charge and electric fields is essential before studying circuits and magnetism.
How do charged objects exert forces on each other across empty space, and how do we quantify the electric field and potential that surround them?
Electrostatics is the study of electric charges at rest. Every proton carries a charge of +e = +1.6 × 10⁻¹⁹ C and every electron carries −e = −1.6 × 10⁻¹⁹ C. The law of conservation of charge states that the net charge of an isolated system never changes — charge can be transferred but never created or destroyed. Objects become charged through friction, conduction, or induction. Once charges are in place, Coulomb's law describes the force between them, the electric field concept extends that force to any test charge, and electric potential gives us an energy-per-charge picture that is essential for understanding circuits and capacitors.
Conductors vs. Insulators & Charging Methods
Conductors
- Electrons move freely throughout the material
- Examples: copper, silver, aluminium, salt water
- Excess charge distributes on the outer surface
- Electric field inside a conductor in equilibrium = 0
Insulators
- Electrons are tightly bound to atoms
- Examples: rubber, glass, plastic, dry wood
- Charge stays localised where it is placed
- Can be polarised by a nearby charge
Charging Methods
Friction
Rubbing two materials transfers electrons from one to the other. Both objects acquire equal and opposite charges (e.g., rubber rod rubbed with fur becomes negative).
Conduction
A charged object touches a neutral conductor; electrons flow until equilibrium. Both objects end up with the same sign of charge.
Induction
A charged object is brought near (no contact). Charges redistribute; if the conductor is then grounded and the charged object removed, the conductor retains the opposite sign.
Key Equations
F = k|q₁q₂| / r²k = 8.99 × 10⁹ N·m²/C²E = F/q = kq/r²N/C or V/m; points away from +V = kq/rVolts (V); scalar quantityU = kq₁q₂/rJoules; negative = attractiveC = Q/VFarads (F)C = ε₀A/dε₀ = 8.85 × 10⁻¹² F/mU = ½CV² = Q²/2CJoulesWorked Examples
Two point charges q₁ = +3 μC and q₂ = −2 μC are separated by r = 0.05 m. Find the magnitude and nature of the electrostatic force between them.
Identify knowns: q₁ = 3 × 10⁻⁶ C, q₂ = 2 × 10⁻⁶ C (magnitudes), r = 0.05 m, k = 8.99 × 10⁹ N·m²/C².
Apply Coulomb's law: F = k|q₁||q₂| / r².
Substitute: F = (8.99 × 10⁹) × (3 × 10⁻⁶) × (2 × 10⁻⁶) / (0.05)².
Numerator: 8.99 × 10⁹ × 6 × 10⁻¹² = 8.99 × 6 × 10⁻³ = 53.94 × 10⁻³ = 0.05394 N·m².
Denominator: (0.05)² = 0.0025 m².
F = 0.05394 / 0.0025 = 21.6 N.
Because the charges are opposite in sign, the force is attractive.
Find the magnitude of the electric field at a point r = 0.3 m from a point charge q = +5 μC.
Identify knowns: q = 5 × 10⁻⁶ C, r = 0.3 m, k = 8.99 × 10⁹ N·m²/C².
Use E = kq / r².
Substitute: E = (8.99 × 10⁹ × 5 × 10⁻⁶) / (0.3)².
Numerator: 8.99 × 10⁹ × 5 × 10⁻⁶ = 44,950 N·m²/C.
Denominator: (0.3)² = 0.09 m².
E = 44,950 / 0.09 = 499,444 N/C ≈ 4.99 × 10⁵ N/C.
Direction: radially outward from the positive charge.
Calculate the electric potential at r = 0.2 m from a point charge q = +8 μC.
Identify knowns: q = 8 × 10⁻⁶ C, r = 0.2 m, k = 8.99 × 10⁹ N·m²/C².
Use V = kq / r.
Substitute: V = (8.99 × 10⁹ × 8 × 10⁻⁶) / 0.2.
Numerator: 8.99 × 10⁹ × 8 × 10⁻⁶ = 71,920 V·m.
V = 71,920 / 0.2 = 359,600 V ≈ 3.60 × 10⁵ V.
Note: potential is a scalar — no direction needed.
A parallel plate capacitor has plate area A = 0.02 m² and plate separation d = 0.001 m. Calculate its capacitance (air gap, ε₀ = 8.85 × 10⁻¹² F/m).
Identify knowns: A = 0.02 m², d = 0.001 m, ε₀ = 8.85 × 10⁻¹² F/m.
Use C = ε₀A / d.
Substitute: C = (8.85 × 10⁻¹² × 0.02) / 0.001.
Numerator: 8.85 × 10⁻¹² × 0.02 = 1.77 × 10⁻¹³ F·m.
C = 1.77 × 10⁻¹³ / 0.001 = 1.77 × 10⁻¹⁰ F = 177 pF.
A capacitor C = 50 μF is charged to a potential difference V = 12 V. Find the charge stored and the energy stored.
Identify knowns: C = 50 × 10⁻⁶ F, V = 12 V.
Charge: Q = CV = 50 × 10⁻⁶ × 12 = 600 × 10⁻⁶ C = 600 μC.
Energy: U = ½CV².
Substitute: U = ½ × 50 × 10⁻⁶ × (12)² = ½ × 50 × 10⁻⁶ × 144.
U = 25 × 10⁻⁶ × 144 = 3,600 × 10⁻⁶ J = 3.6 × 10⁻³ J = 3.6 mJ.
Guided Practice
Two charges q₁ = +6 μC and q₂ = +6 μC are placed 0.10 m apart. Find the electrostatic force between them and state whether it is attractive or repulsive.
Hint: Use F = k|q₁||q₂|/r². Both charges are positive, so the force is repulsive. Remember to square the distance.
A point charge q = −4 μC sits at the origin. What is the electric field magnitude and direction at a point 0.5 m to the right of the charge?
Hint: Use E = k|q|/r². The field points toward a negative charge, so the direction is to the left (toward the charge).
Two charges q₁ = +2 μC at x = 0 and q₂ = −2 μC at x = 0.4 m. Use the superposition principle to find the net electric field at the midpoint x = 0.2 m.
Hint: Calculate E₁ and E₂ separately at the midpoint. Both fields point in the same direction (from + toward −), so add their magnitudes.
A capacitor with C = 20 μF stores a charge Q = 80 μC. Find the potential difference across it and the energy stored.
Hint: First find V = Q/C. Then use U = ½CV² or equivalently U = Q²/(2C).
A parallel plate capacitor has C = 100 pF. A dielectric with κ = 4 is inserted between the plates. What is the new capacitance?
Hint: Inserting a dielectric multiplies capacitance by κ: C_new = κC₀. Multiply the original capacitance by 4.
Key Vocabulary
Electric Charge
A fundamental property of matter (symbol q or Q, unit: coulomb C) that causes electromagnetic interactions. Protons carry +e and electrons carry −e, where e = 1.6 × 10⁻¹⁹ C.
Example: A glass rod rubbed with silk acquires a positive charge because electrons are transferred to the silk.
Coulomb's Law
The electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them: F = k|q₁q₂|/r².
Example: Doubling the distance between two charges reduces the force to one-quarter of its original value.
Electric Field
The force per unit positive test charge at a point in space: E = F/q. For a point charge, E = kq/r². Units: N/C or V/m.
Example: Near a +5 μC charge at 0.3 m, E ≈ 4.99 × 10⁵ N/C directed radially outward.
Electric Potential
The electric potential energy per unit charge at a point: V = U/q = kq/r. It is a scalar measured in volts (V = J/C).
Example: The potential 0.2 m from a +8 μC charge is approximately 3.6 × 10⁵ V.
Potential Difference
The work done per unit charge moving a positive test charge between two points: ΔV = W/q. Also called voltage.
Example: A 9 V battery maintains a 9 V potential difference between its terminals.
Capacitance
The ability of a device to store charge per unit voltage: C = Q/V. Measured in farads (F). For a parallel plate capacitor, C = ε₀A/d.
Example: A 50 μF capacitor charged to 12 V stores 600 μC of charge.
Dielectric
An insulating material placed between capacitor plates that increases capacitance by a factor κ (the dielectric constant): C = κε₀A/d.
Example: Inserting a dielectric with κ = 4 into a 100 pF capacitor increases its capacitance to 400 pF.
Electric Field Lines
Imaginary lines whose tangent at any point gives the direction of the electric field. They originate on positive charges and terminate on negative charges; their density indicates field strength.
Example: Field lines between two parallel plates are parallel and equally spaced, indicating a uniform electric field.
Quick Check — Multiple Choice
Interactive Practice — 5 Questions
Two point charges of +4 μC and −4 μC are separated by 0.2 m. Compared to the original force, what happens to the force if the separation is doubled to 0.4 m?
Which statement correctly describes the electric field inside a conductor in electrostatic equilibrium?
A capacitor stores charge Q when connected to voltage V. If the voltage is tripled while the capacitance stays the same, the energy stored becomes:
Electric field lines between two parallel plates of a capacitor are:
Which charging method results in the charged object and the neutral object acquiring charges of opposite sign?
Independent Practice
Two charges of +3.0 μC and −3.0 μC are placed 0.20 m apart. Calculate the magnitude of the electrostatic force between them. (k = 8.99 × 10⁹ N·m²/C²)
Calculate the electric field strength at a distance of 0.30 m from a point charge of +8.0 μC. (k = 8.99 × 10⁹ N·m²/C²)
Calculate the electric potential at a distance of 0.25 m from a +6.0 μC point charge. (k = 8.99 × 10⁹ N·m²/C²)
An electron is accelerated through a potential difference of 500 V. Find the kinetic energy gained (in joules and eV) and the final speed. (m_e = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
★ A small sphere of mass 2.0 × 10⁻³ kg and charge +4.0 μC is suspended by a string in a uniform horizontal electric field E. The string makes a 30° angle with the vertical. (a) Draw a free-body diagram. (b) Find the electric field strength E. (c) Find the tension in the string. (g = 9.8 m/s²)
ChallengeCommon Mistakes
Confusing electric field E (vector, N/C) with electric potential V (scalar, volts)
E is a vector pointing from high to low potential; V is a scalar (just a number with sign). E = −ΔV/Δr (field points in direction of decreasing potential)
Forgetting that Coulomb's law gives the magnitude of force — direction must be determined separately
F = k|q₁q₂|/r² gives magnitude. Like charges repel (force away from each other); unlike charges attract (force toward each other)
Using superposition by adding force magnitudes instead of vector components
Electric force and field obey vector superposition. Find x- and y-components of each force separately, then add components, then find the resultant magnitude and direction
Thinking a conductor in electrostatic equilibrium has zero electric field everywhere inside it
Inside a conductor in equilibrium: E = 0 and all excess charge resides on the surface. The potential is constant throughout (but not necessarily zero)
Math Tips
Coulomb's constant: k = 8.99×10⁹ N·m²/C² ≈ 9×10⁹. Alternatively, k = 1/(4πε₀) where ε₀ = 8.85×10⁻¹² F/m. Elementary charge: e = 1.6×10⁻¹⁹ C
Electric potential energy vs potential: U = qV. A positive charge moves from high V to low V (like a ball rolling downhill). A negative charge moves from low V to high V
Parallel plate capacitor: E = V/d (uniform field between plates), C = ε₀A/d, Q = CV, U = ½CV² = Q²/2C. With dielectric κ: C_new = κC₀
Capacitors in series: 1/C_total = Σ1/Cᵢ (like resistors in parallel). Capacitors in parallel: C_total = ΣCᵢ (like resistors in series). Remember: series/parallel rules are SWAPPED vs resistors