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Q.

Separation between the plates of a parallel plate capacitor is d and the area of each plate is A . When a slab of material of dielectric constant  ‘k’ and thickness t(t

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a

ε0Ad+t1−1k

b

ε0Ad+t1+1k

c

ε0Ad−t1−1k

d

ε0Ad−t1+1k

answer is C.

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Detailed Solution

Potential difference between the platesV=V1+V2=σε0×d-t+σKε0×t∵E=σε0 &v=Ed ⇒V=σε0d−t+tk =QAε0d−t+1KHence, capacitance C=QV=QQAε0(d−t+tK)=ε0A(d−t+tK)=ε0Ad−t1−1K.
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Separation between the plates of a parallel plate capacitor is d and the area of each plate is A . When a slab of material of dielectric constant  ‘k’ and thickness t(t