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 3/4 d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance ( C 0) 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}

(D) C ^{\prime}=\frac{4}{3+ K } C _{0}

JEE Mains · Physics · STD 12 - 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

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Verified Solution

Correct Answer

(C) \(C ^{\prime}=\frac{4 K }{ K +3} C _{0}\)

Detailed explanation

\(C _{0}=\frac{\epsilon_{0} A }{ d }\)

\(C ^{\prime}= C _{1}\) and \(C _{2}\) in series.

i.e. \(\frac{1}{ C ^{\prime}}=\frac{1}{ C _{1}}+\frac{1}{ C _{2}}\)

\(\frac{1}{ C ^{\prime}}=\frac{(3 d / 4)}{\epsilon_{0} KA }+\frac{ d / 4}{\epsilon_{0} A }\)

\(\frac{1}{ C ^{\prime}}=\frac{ d }{4 \epsilon_{0} A }\left(\frac{3+ K }{ K }\right)\)

\(C ^{\prime}=\frac{4 K C _{0}}{(3+ K )}\)981-s542

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