Study Guides/Physics/Capacitor Charged to V0 Potential — Energy, Charge and Problems
Study Guide · Physics

A Capacitor of Capacitance C Charged to V₀ Potential

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.

Question (Click to Flip)

A capacitor of capacitance C is charged to V₀. What is the energy stored and charge?

Answer

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.

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Key Facts

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).

Key Formulas When Capacitor C is Charged to V₀

  1. Charge stored: Q = CV₀

  2. Energy stored: U = ½CV₀² = Q²/(2C) = QV₀/2

  3. 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)

Questions and Answers

A capacitor of capacitance C is charged to V₀. What is the energy stored and charge?+

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.

A capacitor C charged to V₀ is connected to an identical uncharged capacitor. Find final potential and energy loss.+

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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