Following operations can be performed on a capacitor : $X$ - connect the capacitor to a battery of $emf$ $E.$ $Y$ - disconnect the battery $Z$ - reconnect the battery with polarity reversed. $W$ - insert a dielectric slab in the capacitor
The electric field in the capacitor after the action $ XW$ is the same as that after $ WX$.
The charge appearing on the capacitor is greater after the action $XWY$ than after the action $XYW$.
The electric energy stored in the capacitor is greater after the action $WXY $ than after the action $XYW.$
all of the above
If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \mathrm{~g} / \mathrm{cc}$, the dielectric constant of water will be
(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )
When a dielectric material is introduced between the plates of a charges condenser, then electric field between the plates
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation
The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then