Question

The two plates of a parallel plate capacitor are 4mm apart.A slab of dielectric constant 3 and thickness 3mm is introduced between the plates with its faces parallel to them.the distance between the plates is so adjusted that the capacitance of the capacitor becomes 2/3rd of its original value.what is the new distance between the plates?

Answer 1

Answer:

Explanation:

Here, `d = 4 mm, K = 3` and `t = 3 mm`

Let `d'` be the new distance between the plates.

As `(2)/(3) xx` capacitance with air = Capacitance with dielectric

`:. (2)/(3) (in_(0) A)/(d) = (in_(0) A)/(d' - t + (K)/(2))`

or `(2)/(3) xx (1)/(4) = (1)/(d' - 3 + (3)/(3))`

or `6 = d' - 3 + 1` or `d' = 8 mm`

Answer 2

Explanation:

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