Question

Figure shows two identical capacitors C₁ and C₂, each of 1 μF capacitance connected to a battery of 6 V. Initially switch S is closed. After sometime 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 the switch S is closed, the two capacitors in parallel will be charged by the same potential difference V.

So, charge on capacitor C

q₁ = C₁V = 1 × 6 = 6 mC

q₂ = C₂ V = 1 × 6 = 6 mC

q = q₁ + q₂ = 6 + 6 = 12 mC.

When switch S is opened and dielectric is introduced. Then

Capacity of both the capacitors becomes K times

i.e., C’₁ = C’₂ = KC = 3 × 1 = 3 mF

Capacitor A remains connected to battery

∵ V’₁ = V = 6 V

q’₁ = Kq = 3 x 6 mC = 18 m C

Capacitor B becomes isolated

∵ q’₂ = q₂ or C’₂ V’₂ = C₂V₂ or (KC)V’₂ = CV

V'₂ = v/k = 6/3 =2v