The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K $ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.
the energy stored in the capacitor will become $K$ -times
the force of attraction between the plates will increase to $K^2$-times
the charge on the capacitor will increase to $K$-times
all of the above
Two parallel metal plates having charges $+Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
A parallel - plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :
The respective radii of the two spheres of a spherical condenser are $12\;cm$ and $9\;cm$. The dielectric constant of the medium between them is $ 6$. The capacity of the condenser will be
Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric
A parallel plate capacitor $\mathrm{C}$ with plates of unit area and separation $\mathrm{d}$ is filled with a liquid of dielectric constant $\mathrm{K}=2$. The level of liquid is $\frac{\mathrm{d}}{3}$ initially. Suppose the liquid level decreases at a constant speed $V,$ the time constant as a function of time $t$ is Figure: $Image$