In a parallel plate capacitor of capacitance $C$, a metal sheet is inserted between the plates, parallel to them. If the thickness of the sheet is half of the separation between the plates. The capacitance will be
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A capacitor of $2\,\, \mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2,$ the percentage of its stored energy dissipated is ......$\%$
A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is
Consider a spherical shell of radius $R$ with a total charge $+ Q$ uniformly spread on its surface (centre of the shell lies at the origin $x=0$ ). Two point charges $+q$ and $-q$ are brought, one after the other from far away and placed at $x=-a / 2$ and $x=+a / 2( < R)$, respectively. Magnitude of the work done in this process is
Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.
Reason : Capacity of the capacitor does not depend upon the nature of the material.
Two condensers of capacity $0.3\,\mu F$ and $0.6\,\mu F$ respectively are connected in series. The combination is connected across a potential of $6\,volts$. The ratio of energies stored by the condensers will be
A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\left.\in_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$