A parallel plate capacitor is maintained at a certain potential difference. When a dielectric slab of thickness 3 mm is introduced between the plates, the plate separation had to be increased by 2 mm in order to maintain the same potential difference between the plates. The dielectric constant of the slab is
The capacitance before the introduction of the slab is
If Q is the charge on the plates, the potential difference is ……(1)
Let d' be the new separation between the plates. When a slab of thickness t and dielectric constant K is introduced, the new capacitance is
Since charge Q remains the same, the new potential difference is ……..(2)
Given V' = V. Equating Eqs. (1) and (2), we get
Given d' = d = 2 mm and t = 3 mm. Thus
which gives K = 3.
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The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness d/2 is placed between the plates. The capacitance would then be
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