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
Between the plates of a parallel plate condenser there is $1\,mm$ thick paper of dielectric constant $4$. It is charged at $100\;volt$. The electric field in $volt/metre$ between the plates of the capacitor is
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
A parallel plate condenser has a capacitance $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $'k'$ of the oil is
Two parallel plate capacitors of capacity $C$ and $3\,C$ are connected in parallel combination and charged to a potential difference $18\,V$. The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$. The final potential difference across the combination of capacitors will be $V$
Three capacitors $A,B$ and $C$ are connected with battery $emf\, \varepsilon $. All capacitors are identical initially. If dielectric slab is inserted between plates of capacitor $A$ slowy with help of external force then