Electrostatics Question 130
Question: Capacitance of a parallel plate capacitor becomes 4/3 times its original value if a dielectric slab of thickness t = d/2 is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is [KCET 2003]
Options:
A) 8
B) 4
C) 6
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
$ C _{air}=\frac{{\varepsilon _{0}}A}{d} $ , with dielectric slab C¢= $ \frac{{\varepsilon _{0}}A}{( d-t+\frac{t}{K} )} $
Given $ {C}’=\frac{4}{3}C\Rightarrow $
$ \frac{{\varepsilon _{0}}A}{( d-t+\frac{t}{K} )}=\frac{4}{3}\times \frac{{\varepsilon _{0}}A}{d} $
$ \Rightarrow K=\frac{4t}{4t-d}=\frac{4(d/2)}{4[(d/2)-d]}=2 $
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