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} $



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