Consider a parallel plate capacitor of 10,μ ,F (micro-farad) with air filled in gap between plates.

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2. Electric Potential and Capacitance

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Consider a parallel plate capacitor of $10\,\mu \,F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant $4$, as shown in the figure. The capacity of the capacitor changes to.......$\mu \,F$

$25$

$20$

$40$

$5$

Solution

(a) ${C_1} = \frac{{{\varepsilon _0}\left( {\frac{A}{4}} \right)}}{d},\,{C_2} = \frac{{K{\varepsilon _0}\left( {\frac{A}{2}} \right)}}{d},\,{C_3} = \frac{{{\varepsilon _0}\left( {\frac{A}{4}} \right)}}{d}$

${C_{eq}} = {C_1} + {C_2} + {C_3} = \left( {\frac{{K + 1}}{2}} \right)\frac{{{\varepsilon _0}A}}{d}$$ = \left( {\frac{{4 + 1}}{2}} \right) \times 10 = 25\,\mu F$

Standard 12

Physics

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Consider a parallel plate capacitor of $10\,\mu \,F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant $4$, as shown in the figure. The capacity of the capacitor changes to.......$\mu \,F$

Solution

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