The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of $\in _r = 6$ is inserted in it then energy and capacity becomes (Assume charge on plates remains constant)
$6\,U,\, 6\,C$
$U,\, C$
$\frac{U}{6}\,,6C$
$U,\, 6\,C$
A force $F$ acts between sodium and chlorine ions of salt (sodium chloride) when put $1\,cm$ apart in air. The permittivity of air and dielectric constant of water are ${\varepsilon _0}$ and $K$ respectively. When a piece of salt is put in water electrical force acting between sodium and chlorine ions $1\,cm$ apart is
Write the relation between $\vec P$ and $\vec E$ for a linear isotropic dielectric.
What is dielectric ?
The electric field between the plates of a parallel plate capacitor when connected to a certain battery is ${E_0}$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections, the field between the plates shall be
A parallel plate capacitor with plate area $A$ and plate separation $d =2 \,m$ has a capacitance of $4 \,\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K =3$ (as shown in figure) will be .........$ \mu \,F$