Question

A capacitor has capacitance $5 \mu F$ when it's parallel plates are separated by air medium of thickness d A slab of material of dielectric constant $15$ having area equal to that of plates but thickness $\frac{d}{2}$ is inserted between the plates Capacitance of the capacitor in the presence of slab will be ___$\mu F$

Answer 1

Correct Answer: 6

To determine the capacitance when a dielectric slab is inserted, we use the formula for capacitance with a dielectric: \(C' = \frac{\epsilon \cdot A}{d - t + \frac{t}{k}}\), where \(C'\) is the new capacitance, \(\epsilon\) is the permittivity of free space, \(A\) is the area of the plates, \(d\) is the distance between plates, \(t\) is the thickness of the dielectric, and \(k\) is the dielectric constant.

The correct answer is 6.

=(3d​+2d​)∈0​A​=5d6∈0​A​ 
=56​×5μF=6μF