NEET2015PhysicsCapacitanceActual
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K , which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
Options
- AThe potential difference between the plates decreases K times.
- BThe energy stored in the capacitor decreases K times.
- CThe change in energy stored is 1 2 C V 2 1 K - 1 .
- DThe charge on the capacitor is not conserved.
Correct answer
D. The charge on the capacitor is not conserved.
Step-by-step solution
For air K = 1 , C 1 = C q = C 1 V     … i . The charge remains constant. U 1 = q 2 2 C U 2 = q 2 2 C K Change in energy stored in the capacitor, U = U 2 - U 1 = q 2 2 C   1 K - 1   = 1 2 C V 2   1 K - 1 V ' = q C K ⇒ V ' = V K   Here,1- Potential difference between the plates decreases by K times. 2- The change in energy stored     ∆ U = 1 2   C V 2   1 K - 1 . 3- Charge is conserved. 4- The energy stored in capacitor decrease by K ,