The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value if a dielectric slab of thickness t= $\frac{d}{2}$ is inserted between the plates (where d is the distance of separation between the plates). What is the dielectric constant of the slab?

Question

The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$  times its original value if a dielectric slab of thickness t= $\frac{d}{2}$  is inserted between the plates (where d is the distance of separation between the plates). What is the dielectric constant of the slab?

cdquestions Admin · Accepted Answer

The correct option is (A) : K=2

Answer

Answer

$K=\frac{1}{2}$

Answer

Answer

The capacitance of a parallel plate capacitor becomes \(\frac{4}{3}\) times its original value if a dielectric slab of thickness t=\(\frac{d}{2}\) is inserted between the plates (where d is the distance of separation between the plates). What is the dielectric constant of the slab?

The Correct Option is A

Solution and Explanation

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