Neet Solved Paper 2015 Question 26
Question: 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:
A) The potential difference between the plates decreases K times
B) The energy stored in the capacitor decreases K times
C) The change in energy stored is $ \frac{1}{2}CV^{2}( \frac{1}{K}-1 ) $
D) The charge on the capacitor is not conserved
Show Answer
Answer:
Correct Answer: D
Solution:
- When a parallel plate air capacitor connected to a cell of emf V, then charge stored will be $ q=CV $
$ \Rightarrow ,V=\frac{q}{C} $ Also energy stored is $ U=\frac{1}{2}CV^{2}=\frac{q^{2}}{2C} $ As the battery is disconnected from the capacitor the charge will not be destroyed i.e. $ q’=q $ with the introduction of dielectric in the gap of capacitor the new capacitance will be $ C’=CK $
$ \Rightarrow $ $ V’=\frac{q}{C’}=\frac{q}{CK} $ The new energy stored will be $ U’=\frac{q^{2}}{2CK} $ $ \Delta U=U’-U=\frac{q^{2}}{2C}( \frac{1}{K}-1 ) $ $ =\frac{1}{2}CV^{2}( \frac{1}{K}-1 ) $ Hence, option (d) is Incorrect.
