Electrostatics Question 816
Question: A combination of parallel plate capacitors is maintained at a certain potential difference. When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.
Options:
3
4
5
6
Show Answer
Answer:
Correct Answer: C
Solution:
Before introducing a slab capacitance of plates $ C _{1}=\frac{{\varepsilon _{0}}A}{d} $ If a slab of dielectric constant K is introduced between plates then $ C=\frac{K{\varepsilon _{0}}A}{d}\text{ then }C _{1}’=\frac{{\varepsilon _{0}}A}{2.4} $
$ C _{1} $ and $ C _{1}’ $ are in series hence, $ \frac{{\varepsilon {0}}A}{d{1}}=\frac{k\frac{{\varepsilon {0}}A}{d{1}}.\frac{{\varepsilon {0}}A}{d{2}}}{k\frac{{\varepsilon {0}}A}{d{1}}+\frac{{\varepsilon {0}}A}{d{2}}} $
$ 3k=2.4k+3\text{ }0.6k=3 $ Hence, the dielectric constant of slab is given by, $ k=\frac{30}{6}=5 $
BETA
