JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Two metallic plates form a parallel plate capacitor. The distance between the plates is \('d'.\) A metal sheet of thickness \(\frac{d}{2}\) and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor \(?\)
- A \(2: 1\)
- B \(1: 2\)
- C \(1: 4\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(A) \(2: 1\)
Step-by-step Solution
Detailed explanation
\(C _{1}=\frac{\epsilon_{0} A }{ d }\) \(C _{2}=\frac{\epsilon_{0} A }{\frac{ d }{2}+\frac{ d / 2}{\propto}}=\frac{2 \epsilon_{0} A }{ d }\) \(\frac{ C _{2}}{ C _{1}}=\frac{2}{1}\)
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