A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is
$K$
$1/K$
$\frac{A}{{{d^2}K}}$
$\frac{{{d^2}K}}{A}$
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
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
The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used, is $C$. If the dielectric is replaced by another with dielectric constant $20$, the capacity will become
Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be
A parallel palate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be