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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

  1. A \(C ^{\prime}=\frac{3+ K }{4 K } C _{0}\)
  2. B \(C ^{\prime}=\frac{4+ K }{3} C _{0}\)
  3. C \(C ^{\prime}=\frac{4 K }{ K +3} C _{0}\)
  4. D \(C ^{\prime}=\frac{4}{3+ K } C _{0}\)
Verified Solution

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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