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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
The distance between two plates of a capacitor is \(d\) and its capacitance is \(C _1\), when air is the medium between the plates. If a metal sheet of thickness \(\frac{2 d }{3}\) and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes \(C _2\). The ratio \(\frac{ C _2}{ C _1}\) is:
- A \(2: 1\)
- B \(4: 1\)
- C \(3: 1\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(C) \(3: 1\)
Step-by-step Solution
Detailed explanation
\(K _{\text {metal sheet }}=\infty, t =\frac{2 d }{3}\) \(C _1=\frac{\in_0 A }{ d }\) \(C _2=\frac{\in_0 A }{ d - t +\frac{ t }{ k }}=\frac{\in_0 A }{ d -\frac{2 d }{3}+0}=3 C _1\) \(\frac{ C _2}{ C _1}=3\)
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