Electrostatics Question 644
Question: The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $ C _{0} $ and $ W _{0} $ . If the air is replaced by glass (dielectric constant = 5) between the plates, the capacity of the plates and the energy stored in it will respectively be
Options:
A) $ 5{C _{0,}}5W _{0} $
B) $ 5{C _{0,}}\frac{W _{0}}{5} $
C) $ \frac{C _{0}}{5},5W _{0} $
D) $ \frac{C _{0}}{5},\frac{W _{0}}{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] When a dielectric K is introduced in a parallel plate condenser its capacity becomes K times.
Hence $ C’=5C _{0} $
Energy stored $ W _{0}=\frac{q^{2}}{2C _{0}} $
$ \therefore W’=\frac{q^{2}}{2C’}=\frac{q^{2}}{2\times 5C _{0}}\Rightarrow W’=\frac{W _{0}}{5} $
