Electrostatics Question 13

Question: 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 constantk and thickness t(t<d) is introduced between the plates, its capacitance becomes [MP PMT 1989]

Options:

A) ε0Ad+t(11k)

B) ε0Ad+t(1+1k)

C) ε0Adt(11k)

D) ε0Adt(1+1k)

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Answer:

Correct Answer: C

Solution:

Potential difference between the plates V=Vair+Vmedium

=σε0×(dt)+σKε0×t

therefore V=σε0(dt+tK)

=QAε0(dt+tK)

Hence capacitance C=QV=QQAε0(dt+tK)

=ε0A(dt+tk)=ε0Adt(11k)



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