SATHEE: Electrostatics Question 816

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:

A) 3

B) 4

C) 5

D) 6

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Before introducing a slab capacitance of plates $C_{1} = \frac{ε_{0} A}{3}$ If a slab of dielectric constant K is introduced between plates then $C = \frac{K ε_{0} A}{d} then C_{1}^{'} = \frac{ε_{0} A}{2.4}$

$C_{1}$ and $C_{1}^{'}$ are in series hence, $\frac{ε_{0} A}{3} = \frac{k \frac{ε_{0} A}{3} . \frac{ε_{0} A}{2.4}}{k \frac{ε_{0} A}{3} + \frac{ε_{0} A}{2.4}}$

$3 k = 2.4 k + 3 0.6 k = 3$ Hence, the dielectric constant of slap in given by, $k = \frac{30}{6} = 5$

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Electrostatics Question 816

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