Consider a parallel plate capacitor of $10\,\mu \,F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant $4$, as shown in the figure. The capacity of the capacitor changes to.......$\mu \,F$
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Change $Q$ on a capacitor varies with voltage $V$ as shown in the figure, where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents
Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is $2 \times {10^5}\,V/m$. When the space is filled with dielectric, the electric field becomes $1 \times {10^5}\,V/m$. The dielectric constant of the dielectric material
A parallel plate capacitor is charged and then disconnected from the source steady $E.M.F.$ The plates are then drawn apart farther. Again it is connected to the same source. Then :
If the dielectric constant and dielectric strength be denoted by $k$ and $x$ respectively, then a material suitable for use as a dielectric in a capacitor must have
A parallel plate capacitor with plates of area $1\,m^2$ each, are at a separation of $0.1\,m.$ If the electric field between the plates is $100\,N/C,$ the magnitude of charge on each plate
A parallel plate capacitor has a plate separation of $0.01\, mm$ and use a dielectric (whose dielectric strength is $19\, KV/mm$) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is......$V$