Question
In the figure a capacitor is filled with dielectrics. The resultant capacitance is
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Answer
(d) ${C_1} = \frac{{{K_1}{\varepsilon _0}\frac{A}{2}}}{{\left( {\frac{d}{2}} \right)}} = \frac{{{K_1}{\varepsilon _0}A}}{d}$
${C_2} = \frac{{{K_2}{\varepsilon _0}\left( {\frac{A}{2}} \right)}}{{\left( {\frac{d}{2}} \right)}} = \frac{{{K_2}{\varepsilon _0}A}}{d}$and ${C_3} = \frac{{{K_3}{\varepsilon _0}A}}{{2d}} = \frac{{{K_3}{\varepsilon _0}A}}{{2d}}$
Now, ${C_{eq}} = {C_3} + \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$$ = \left( {\frac{{{K_3}}}{2} + \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right).\frac{{{\varepsilon _0}A}}{d}$
${C_2} = \frac{{{K_2}{\varepsilon _0}\left( {\frac{A}{2}} \right)}}{{\left( {\frac{d}{2}} \right)}} = \frac{{{K_2}{\varepsilon _0}A}}{d}$and ${C_3} = \frac{{{K_3}{\varepsilon _0}A}}{{2d}} = \frac{{{K_3}{\varepsilon _0}A}}{{2d}}$
Now, ${C_{eq}} = {C_3} + \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$$ = \left( {\frac{{{K_3}}}{2} + \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right).\frac{{{\varepsilon _0}A}}{d}$
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