A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?.....$\mu F$
$96$
$24$
$50$
$10$
A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect $?$
The distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $ x=0$ and negatively charged plate is at $x=5d$. Two slabs one of conductor and other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like :
Explain polarisation of polar molecule in uniform electric field.
Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.
Reason : Capacity of the capacitor does not depend upon the nature of the material.
Eight small drops, each of radius $r$ and having same charge $q$ are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is