JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor has plate of length \('l',\) width \('w'\) and separation of plates is \('d'.\) It is connected to a battery of emf \(V\). A dielectric slab of the same thickness '\(d\)' and of dielectric constant \(k =4\) is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored\(?\)
- A \(\frac{l}{4}\)
- B \(\frac{l}{2}\)
- C \(\frac{l}{3}\)
- D \(\frac{2l}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{l}{3}\)
Step-by-step Solution
Detailed explanation
Before inserting slab \(C_{i}=\frac{\varepsilon_{0} A}{d}\) \(C_{i}=\frac{\varepsilon_{0}(w)}{d}\) After inserting dielectric slab \(C_{f}=C_{1}+C_{2}\) \(C_{f}=\frac{K \varepsilon_{0} A_{1}}{d}+\frac{\varepsilon_{0} A_{2}}{d}\)…
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