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Class 12  »  Physics  »  Electrostatic Potential and Capacitance

Energy Stored in a Capacitor - UNSOLVED PRACTICE SET

Class 12

Chapter: Electrostatic Potential and Capacitance | Topic: Energy Stored in a Capacitor

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Study Material.
Class 12

ENERGY STORED IN A CAPACITOR - UNSOLVED PRACTICE SET

Topic: Energy Stored in a Capacitor

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. The energy stored in a charged capacitor is given by:

  1. ½QV
  2. ½CV²
  3. Q²/2C
  4. All of the above

Q2. The energy density (energy per unit volume) in the electric field between the plates of a parallel plate capacitor is:

  1. ½ε₀E
  2. ½ε₀E²
  3. ε₀E²
  4. ½E²/ε₀

Q3. When a capacitor is connected to a battery and then the battery is disconnected, the energy stored in the capacitor:

  1. Becomes zero
  2. Remains the same if plate separation is unchanged
  3. Increases if plate separation is increased
  4. Decreases if plate separation is increased

Q4. The energy stored in a capacitor is stored in:

  1. The conducting plates
  2. The dielectric between the plates
  3. The electric field between the plates
  4. The battery connected to it

Q5. If the potential difference across a capacitor is doubled (with battery connected), the energy stored becomes:

  1. Half
  2. Same
  3. Double
  4. Four times

Q6. Two identical capacitors, each with capacitance C, are connected in parallel to a battery of voltage V. The total energy stored is:

  1. ½CV²
  2. CV²
  3. 2CV²
  4. ¼CV²

Short Answer Questions

Q7. Derive the expression for the energy stored in a capacitor in terms of Q, C, and V.

Q8. A capacitor of capacitance 10 μF is charged to 100 V. Calculate the energy stored in it.

Q9. Why does the energy stored in a capacitor increase when the plate separation is increased after disconnecting the battery? Where does this extra energy come from?

Q10. Define energy density in an electric field. Write its expression for a parallel plate capacitor.

Q11. A 2 μF capacitor is charged to 50 V and then connected in parallel with an uncharged 4 μF capacitor. What happens to the total energy of the system? Explain.

Q12. The energy stored in a capacitor is 0.5 J when charged to 100 V. What is its capacitance?

Long Answer Questions

Q13. Derive the expression for the energy stored in a charged capacitor. Show that it can be expressed in three equivalent forms: U = ½QV = ½CV² = Q²/2C. Explain the physical significance of each form.

Q14. Derive the expression for the energy density in the electric field of a parallel plate capacitor. Show that it is equal to ½ε₀E².

Q15. A parallel plate capacitor has capacitance C = 5 μF. It is charged to a potential difference V = 200 V and then isolated from the battery.

(a) Calculate the charge, energy stored, and energy density.

(b) The plate separation is now doubled. Calculate the new capacitance, potential difference, charge, and energy.

(c) Account for the change in energy. Where did the extra energy come from?

Numerical / Application-Based Problems

Q16. A parallel plate capacitor has plate area A = 200 cm² and plate separation d = 1 mm. It is connected to a 100 V battery.

(a) Calculate the capacitance, charge, and energy stored.

(b) The battery is disconnected and a dielectric slab of K = 5 and thickness 1 mm is inserted. Calculate the new charge, potential difference, and energy.

(c) The dielectric is now removed and the battery is reconnected. Calculate the new charge and energy.

(d) Compare the energy in parts (a), (b), and (c) and explain the differences.

[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²]

Q17. In a hospital, a defibrillator uses a capacitor to deliver a life-saving electric shock to a patient's heart. A typical defibrillator capacitor has C = 100 μF and is charged to V = 2000 V.

(a) Calculate the energy stored in the capacitor.

(b) If this energy is delivered in 5 milliseconds, what is the average power delivered?

(c) The shock delivers about 200 J of energy to the patient. What is the efficiency of the defibrillator?

(d) Why is a capacitor used instead of a battery for this application?

Q18. Two capacitors C₁ = 4 μF and C₂ = 6 μF are connected in series across a 100 V battery.

(a) Calculate the equivalent capacitance and the total energy stored.

(b) Find the charge on each capacitor and the energy stored in each.

(c) The capacitors are now disconnected from the battery and connected in parallel (positive plate to positive plate). Calculate the common potential difference and the total energy.

(d) Account for the loss of energy, if any. Where did it go?


Total: 30 Marks | Time: 40 mins

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