Questions
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 is introduced between the plates its capacitance becomes
detailed solution
Correct option is C
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.Talk to our academic expert!
Similar Questions
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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