NEET · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor having crosssectional area \(A\) and separation \(d\) has air in between the plates. Now an insulating slab of same area but thickness \(d/2\) is inserted between the plates as shown in figure having dielectric constant \(K (=4) .\) The ratio of new capacitance to its original capacitance will be,
- A \(4:1\)
- B \(2:1\)
- C \(8:5\)
- D \(6:5\)
Answer & Solution
Correct Answer
(C) \(8:5\)
Step-by-step Solution
Detailed explanation
\(C_{0}=\frac{\varepsilon_{0} A}{d}\) After inserting dielectric
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