Capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$, dielectric constant of paper is $2.5$ and width of paper is $400\,mm$, then length of foil will be.....$m$
$0.34$
$1.33$
$13.4$
$33.9$
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 of plate area $A$ and plate separation $d$ is charged to potential $V$ and then the battery is disconnected. A slab of dielectric constant $k$ is then inserted between the plates of the capacitors so as to fill the space between the plates. If $Q,\;E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then state incorrect relation from the following
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )
Explain polarization of nonpolar molecule in uniform electric field and define the linear isotropic dielectrics.
The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$