$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

Two square-shaped metal plates of side $1 \,m$, kept $0.01 \,m$ apart in air, form a parallel plate capacitor. It is connected to a battery of $500 \,V$. The plates of the capacitor are then immersed in an insulating oil by lowering the plates vertically with a speed of $0.001 \,m/s$. If the dielectric constant of the oil is $11$, then the current drawn from the battery during this process is:

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

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 the voltage source remaining connected to the capacitor.

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

$A$ parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles,then