Space between the plates of a parallel plate capacitor of plate area 4 and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 F is connected

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Solution

Q.
Space between the plates of a parallel plate capacitor of plate area 4 ${cm}^{2}$ and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is ______ pF. (Given $ε_{0} = 8.85 \times 10^{- 12} F / m$ ) [2025]

Ans.
(15)

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{d / 2}{A k_{1} ε_{0}} + \frac{d / 2}{A k_{2} ε_{0}}$

$= (\frac{1}{k_{1}} + \frac{1}{k_{2}}) \frac{d}{2 A ε_{0}} = (\frac{1}{3} + \frac{1}{5}) \frac{d}{2 A ε_{0}}$

$\frac{1}{C} = \frac{4}{15} \frac{d}{A ε_{0}}$

$C = \frac{15}{4} \times \frac{A ε_{0}}{d}$

$= \frac{15}{4} \times \frac{4 \times 10^{- 4} \times 8.85 \times 10^{- 12}}{1.77 \times 10^{- 3}} = 7.5 p F$

Finally equivalent capacitance ( $C_{eq}$ ) = 7.5 + 7.5 = 15 pF

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