When a capacitor of capacitance C is charged to a potential V₀, it stores charge Q = CV₀ and energy U = ½CV₀². This is a fundamental concept in Class 12 Physics (Electrostatics / Capacitors). Problems based on this setup often involve energy stored, charge redistribution when two capacitors are connected, and energy loss.
Charge on capacitor: Q = CV₀.
Energy stored: U = ½CV₀² = Q²/(2C).
Two equal capacitors (C each): if one charged to V₀ and connected to uncharged one, final V = V₀/2.
Energy loss when two capacitors connected = ¼CV₀² (dissipated as heat).
If disconnected and plates doubled: V doubles to 2V₀; energy doubles to CV₀².
Energy can INCREASE when capacitor is disconnected and plates are separated (work done by external agent).
Charge stored: Q = CV₀
Energy stored: U = ½CV₀² = Q²/(2C) = QV₀/2
Electric field (for parallel plate capacitor, plate separation d): E = V₀/d
Common Problem Type 1 — Two Capacitors (Charge Redistribution): 'A capacitor C₁ is charged to V₀. It is then connected in parallel with an uncharged capacitor C₂. Find the final potential and energy loss.'
Solution: • Initial charge: Q = C₁V₀ • Charge is conserved: Q = (C₁ + C₂) × V_final • Final potential: V_f = C₁V₀ / (C₁ + C₂)
Special case: C₁ = C₂ = C: • V_f = C × V₀ / (2C) = V₀/2 • Final potential = V₀/2
Energy Loss: • Initial energy: U_i = ½C₁V₀² • Final energy: U_f = ½(C₁ + C₂)V_f² • For C₁ = C₂ = C: U_f = ½(2C)(V₀/2)² = ¼CV₀² • Energy loss = ½CV₀² − ¼CV₀² = ¼CV₀² • This energy is dissipated as heat in the connecting wire.
Common Problem Type 2 — Capacitor Disconnected, then Separation Changed: 'A parallel plate capacitor C is charged to V₀ and then disconnected from battery. The plate separation is doubled. Find the new potential and energy.'
Solution: • Charge Q = CV₀ remains constant (disconnected from battery) • New capacitance C' = C/2 (since C ∝ 1/d) • New potential V' = Q/C' = CV₀/(C/2) = 2V₀ • New energy U' = Q²/(2C') = (CV₀)²/(2 × C/2) = CV₀² • Initial energy = ½CV₀²; New energy = CV₀² • Energy increased (work done by external force pulling plates apart)
Charge stored: Q = CV₀. Energy stored: U = ½CV₀². These are the fundamental relations for a charged capacitor. The energy can also be written as U = Q²/(2C) = QV₀/2.
Final potential = V₀/2 (charge Q = CV₀ is shared equally between 2 capacitors of equal capacitance). Initial energy = ½CV₀². Final energy = ½(2C)(V₀/2)² = ¼CV₀². Energy lost = ½CV₀² − ¼CV₀² = ¼CV₀². This energy is dissipated as heat.
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