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Question

An isolated parallel plate capacitor is charged upto a certain potential difference. When a 3mm thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by 2.4mm. Find the dielectric constant of the slab. (Assume charge remains constant)

A
5
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B
10
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C
2.5
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D
7.5
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Solution

The correct option is A 5
Cf=ε0Adt(11K)
Ci=Cf
ε0Ad=ε0Adt(11K)d=dt(11K)
d=d2.4×103
t=3mm
d=d+(24×103)3×103(11K)
K=5

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