NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
A certain series R-C circuit is formed using resistance R, a capacitor without dielectric having a capacitance C = 2 F and a battery of emf E = 3 V. The circuit is completed and it is allowed to attain the steady state. After this, at t = 0 half the thickness of the capacitor is filled with a dielectric constant k = 2 as shown in the figure. The system is again allowed to attain a steady state. What will be the heat
Options
- A3
- B5
- C2
- D6
Correct answer
A. 3
Step-by-step solution
Initial charge on capacitor = CE Initial potential energy of capacitor = CE 2 /2 Now, C = ε 0 A/d And new capacitance C ′ = ε 0 A d/2 , in series with = ε 0 K A d/2 ⇒ C ' = 2 ∈ 0 A d in series with 4 ∈ 0 A d = ε 0 A d 2 × 4 2 + 4 = 4 ε 0 A 3 d = 4 3 C ⇒ New charge = 4 3 CE New energy = 1 2 × 4 3 CE 2 = 2 3 CE 2 Now W battery = Δ H + Δ U ⇒ E 4 3 CE - CE = Δ H + 2 3 CE 2 - 1 2 CE 2 ⇒ Δ H = 1 6 CE 2 = 3