Question

Figure shows two identical capacitors, C₁ and C₂, each of 1 mF capacitance connected to a battery of 6 V. Initially switch ‘S’ is closed. After sometimes ‘S’ is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the

(i) charge and

(ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Answer 1

When switch S is closed, p.d. across each capacitor is 6V

V₁ = V₂ = 6 V

C₁ = C₂ = 1 µC

∴ Charge on each capacitor

q₁ = q₂ (= CV) = (1 µF) × (6 V) = 6 µC

When switch S is opened, the p.d. across C₁ remains 6 V, while the charge on capacitor C₂ remains 6 µC. After insertion of dielectric between the plates of each capacitor, the new capacitance of each capacitor becomes

C′₁ = C′₂ = 3 × 1 µF = 3 µF

Charge on capacitor C₁, q′1 = C′₁ V₁ = (3 µF) × 6 V = 18 µC Charge on capacitor C₂ remains 6 µ

Potential difference across C₁ remains 6 V. Potential difference across C₂ becomes