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 $\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$
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The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it, one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the condenser is
In an insulated parallel-plate capacitor of capacitance $C$, the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plate is
The charge given to a hollow sphere of radius $10\, cm$ is $3.2×10^{-19}\, coulomb$. At a distance of $4\, cm$ from its centre, the electric potential will be
Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is........$times$
The diagram shows three infinitely long uniform line charges placed on the $X, Y $ and $Z$ axis. The work done in moving a unit positive charge from $(1, 1, 1) $ to $(0, 1, 1) $ is equal to
The four identical capacitors in the circuit shown below are initially uncharged. The switch is then thrown first to position $A$, and then to position $B$. After this is done:
Note: $V_{1,2,3,4}$ are the potential differences across $C_{1,2,3,4}$ and $Q_{1,2,3,4}$ are the final charges stored in $C_{1,2,3,4}$ respectively.