Initial energy (only C₁ is charged, C₂ is uncharged):
Uᵢ = ½CV² = ½ × 200×10⁻¹² × (100)² = ½ × 200×10⁻¹² × 10⁴
Uᵢ = 1·0 × 10⁻⁶ J
Common voltage after connection (charge is conserved):
Total charge Q = CV = 200×10⁻¹² × 100 = 2×10⁻⁸ C
Total capacitance = C + C = 2C = 400 pF
V' = Q/(2C) = CV/(2C) = V/2 = 50 V
Final energy (both capacitors at 50 V):
Uf = ½(2C)V'² = ½ × 400×10⁻¹² × (50)²
Uf = ½ × 400×10⁻¹² × 2500 = 0·5 × 10⁻⁶ J
Energy lost:
ΔU = Uᵢ − Uf = 1·0×10⁻⁶ − 0·5×10⁻⁶ = 0·5 × 10⁻⁶ J
ΔU = ½CV² − ½(2C)(V/2)² = ½CV² − ¼CV² = ¼CV²
= ¼ × 200×10⁻¹² × 10⁴ = 0·5 × 10⁻⁶ J
Energy lost = HALF the initial energy (always, for identical capacitors)
A capacitor stores electrical energy in the electric field between its plates. The energy stored is: U = ½CV² = Q²/(2C) = QV/2. All three forms are equivalent (using Q = CV). For this problem: U = ½ × 200×10⁻¹² × (100)² = 10⁻⁶ J = 1 μJ. This energy comes from the work done by the battery in moving charge from one plate to the other, against the growing electric field.
When a charged capacitor (V) is connected to an uncharged identical capacitor, charge flows until both reach the same potential V/2. Charge is conserved: Q_initial = Q_final. But energy is NOT conserved — half is lost. Where does it go? To heat in the connecting wires (even if resistance is zero, there is still a spark — electromagnetic radiation is emitted). This is an unavoidable consequence of the discontinuous redistribution of charge. It's a fundamental result: connecting two identical capacitors always wastes exactly 50% of the initial energy, regardless of the capacitance value.
📌 Charge is always conserved when capacitors are connected
📌 Total charge before = Total charge after: Q = CV (on C₁) + 0 (on C₂) = CV
📌 After connection: Q₁' + Q₂' = Q₁ + Q₂ → still = CV
📌 Energy is NOT conserved — half is lost to heat/radiation
📌 Final energy = ½ × initial energy (for identical capacitors)
📌 The "missing" energy appears as heat in wires or EM radiation
Series combination: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + .... Same charge Q on each capacitor; voltages add: V = V₁ + V₂ + .... Used when need higher voltage rating. Parallel combination: C_eq = C₁ + C₂ + C₃ + .... Same voltage V across each; charges add: Q = Q₁ + Q₂ + .... Used when need higher capacitance. In this problem: when the two capacitors are connected (positive plate to positive, negative to negative), they are effectively in parallel: C_eq = C₁ + C₂ = 2C.
Inserting a dielectric (insulating material) between capacitor plates increases capacitance: C = κC₀ = κε₀A/d, where κ is the dielectric constant (κ ≥ 1). This happens because the dielectric gets polarised — induced surface charges partially cancel the plate charges, reducing the effective electric field, allowing more charge to be stored at the same voltage. Effect on a charged capacitor: (1) Connected to battery: charge increases (Q = κCV), energy increases, electric field stays same. (2) Disconnected from battery: voltage decreases, energy decreases (energy used to pull in the dielectric).
📌 Capacitance: C = ε₀A/d (in vacuum), C = κε₀A/d (with dielectric)
📌 Electric field between plates: E = σ/ε₀ = V/d (uniform)
📌 Energy density in electric field: u = ½ε₀E² J/m³
📌 Total energy: U = u × Volume = ½ε₀E² × Ad = ½CV²
📌 Force between plates: F = Q²/(2ε₀A) = ½ε₀E²A (attractive)
📌 Capacitance increases if: A increases, d decreases, κ increases
Capacitors are used everywhere in electronics. Electrolytic capacitors (large C, 1μF–10,000μF): used for power supply filtering, audio coupling. Ceramic capacitors (small C, 1pF–1μF): used in high-frequency circuits. Variable capacitors: used in radio tuning. Supercapacitors (up to farads!): energy storage in hybrid vehicles, backup power. Camera flash: capacitor stores charge, then dumps it quickly through the flash bulb — much higher instantaneous power than battery alone can provide.
Both capacitors and batteries store energy, but very differently. Capacitors: store energy in electric fields, charge/discharge extremely fast (microseconds to milliseconds), lower energy density (Wh/kg) but higher power density (W/kg). Batteries: store energy in chemical bonds, slow to charge (hours), higher energy density but lower power density. Supercapacitors fill the gap. A 200 pF capacitor at 100V stores only 1 μJ — a typical AA battery stores ~3000 Wh × 10⁻⁶ = 3 Wh = 10,800 J — about 10 billion times more energy!