JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant \(K\) is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is \(\frac{3}{4} d\), where \('d'\) is the separation between the plates of parallel plate capacitor. The new capacitance \((C')\) in terms of original capacitance \(\left( C _{0}\right)\) is given by the following relation
- A \(C ^{\prime}=\frac{3+ K }{4 K } C _{0}\)
- B \(C ^{\prime}=\frac{4+ K }{3} C _{0}\)
- C \(C ^{\prime}=\frac{4 K }{ K +3} C _{0}\)
- D \(C ^{\prime}=\frac{4}{3+ K } C _{0}\)
Answer & Solution
Correct Answer
(C) \(C ^{\prime}=\frac{4 K }{ K +3} C _{0}\)
Step-by-step Solution
Detailed explanation
\(C _{0}=\frac{\epsilon_{0} A }{ d }\) \(C ^{\prime}= C _{1}\) and \(C _{2}\) in series. i.e. \(\frac{1}{ C ^{\prime}}=\frac{1}{ C _{1}}+\frac{1}{ C _{2}}\) \(\frac{1}{ C ^{\prime}}=\frac{(3 d / 4)}{\epsilon_{0} KA }+\frac{ d / 4}{\epsilon_{0} A }\)…
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