$A$ parallel plate capacitor whose capacitance $C$ is $14 \, pF$ is charged by a battery to a potential difference $V = 12 \, V$ between its plates. The charging battery is now disconnected and a porcelain plate with dielectric constant $k = 7$ is inserted between the plates. The plate would oscillate back and forth between the plates with a constant mechanical energy of $.......... pJ$. (Assume no friction)
- A$872$
- B$972$
- C$784$
- D$864$
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View Solution$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \text{ pF}$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4 \text{ pF}$. The value of $x$ is
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View Solution$A$ parallel plate capacitor of capacitance $12.5 \ pF$ is charged by a battery connected between its plates to a potential difference of $12.0 \ V$. The battery is now disconnected and a dielectric slab $(\epsilon_{r}=6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is . . . . . . . $\times 10^{-12} \ J$.
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View SolutionAn air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).
$A$ parallel plate capacitor has a capacitance $C$ and the distance between the plates is $d$. If the space between the plates is filled with a dielectric material of dielectric constant $K$ as shown in the figure,find the new capacitance of the capacitor.
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