$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)

  • A
    $10$
  • B
    $20$
  • C
    $30$
  • D
    $40$

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Similar Questions

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

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In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. The capacitance of the capacitor is $17.71 \, pF$. If this capacitor is connected to a $100 \, V$ supply,and a $3 \, mm$ thick mica sheet (of dielectric constant $k = 6$) is inserted between the plates,calculate the new capacitance,charge,and potential difference in the following cases:
$(a)$ While the voltage supply remains connected.
$(b)$ After the supply is disconnected.

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