Capacitors & Capacitance — A/L Physics

How do I practise Capacitors & Capacitance?

1 question with a worked solution

1. A parallel-plate capacitor has capacitance C = 10 μF with air between its plates and is connected across a 12 V battery. If a dielectric slab of relative permittivity ε_r = 5 is inserted to completely fill the space between the plates while the battery remains connected, what is the new electrostatic energy stored in the capacitor?
   1. A) 0.72 mJ
   2. B) 1.8 mJ
   3. C) 7.2 mJ
   4. D) 3.6 mJ
   Check my answer

   Correct answer: D) 3.6 mJ

   When a dielectric slab of dielectric constant ε_r is inserted, capacitance increases: C' = ε_r C = 5 × 10 μF = 50 μF. Because the battery remains connected, the voltage V remains constant at 12 V. New stored energy U' = ½ C' V² = ½ (50 × 10⁻⁶ F) × (12 V)² = 25 × 10⁻⁶ × 144 = 3600 × 10⁻⁶ J = 3.6 × 10⁻³ J = 3.6 mJ.

   Show worked solution

   Step 1: New capacitance C' = ε_r · C = 5 × 10 μF = 50 μF. Step 2: Voltage V = 12 V (constant since battery connected). Step 3: U' = ½ C' V² = ½ (50 × 10⁻⁶)(144) = 3.6 × 10⁻³ J = 3.6 mJ. Answer: 3.6 mJ.
   Modelled on: the A/L Physics examination pattern

Get this auto-marked

Next in the syllabus

Where does Capacitors & Capacitance fit in the syllabus?

Unit 6.5 · Year 1 · Term 3