# Joint Entrance Examination/2000 Screening/Q023 capcitors and dielectric, equivalent capcitance

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# Original Question by Dharav talk

Here's the question...
 A parallel plate capaciter of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k1, k2 and k3 as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor then its dielectric constant k is given by... ${\displaystyle {\frac {1}{k}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}+{\frac {1}{2k_{3}}}}$ ${\displaystyle {\frac {1}{k}}={\frac {1}{k_{1}+k_{2}}}+{\frac {1}{2k_{3}}}}$ ${\displaystyle {\frac {1}{k}}={\frac {k_{1}k_{2}}{k_{1}+k_{2}}}+{\frac {1}{k_{2}}}+{2k_{3}}}$ ${\displaystyle {\frac {1}{k}}={\frac {k_{3}k_{1}}{k_{1}+k_{2}}}+{\frac {k_{2}k_{3}}{k_{2}+k_{3}}}}$ Posted by Dharav talk 05:08, 18 August 2009 (UTC)

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