Electrostatic P...

The capacitance of a capacitor becomes $$\dfrac{7}{6}$$ times its original value if a dielectric slab of thickness, $$t=\dfrac{2}{3}d$$ is introduced in between the plates. d is the separation between the plates. The dielectric constant of the dielectric slab is :       

In the capacitor of capacitance $$20\  \mu F$$, the distance between plates is $$2\ mm$$. If a material of dielectric constant $$ 2$$ is inserted between the plates, then the capacitance of the system is :

A highly conducting sheet of aluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates at distance $$\dfrac{d}{2}$$ from positive plate where $$d$$ is distance between plates. If the capacitance before the insertion of foil was $$10 \; \mu F$$ , its value after the insertion of foil will be:

Two metal plates are separated by a distance $$d$$ in a parallel plate condenser. A metal plate of thickness $$t$$ and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of : 

An oil condenser has a capacity of $$100\; \mu F$$ . The oil has dielectric constant 2. When the oil leaks out , its new capacity is :

A positive point charge q is carried from a point B to a point A in the electric field of a point charge +Q. If the permittivity of free space is $$\epsilon _{0}$$ , the work done in the process is given by

A capacitor of $$10 \mu F$$ capacitance is charged by a $$12 V$$ battery. Now the space between the plates of capacitors is filled with a dielectric of dielectric constant $$K = 3$$ and again it is charged. The magnitude of the charge is :

The capacity of a parallel plate condenser is $$10 \;\mu F$$ without the dielectric. Material with a dielectric constant of $$2$$ is used to fill half-thickness between the plates. The new capacitance is $$\underline{\hspace{0.5in}} \;\mu F$$ :

The plates of a parallel plate capacitor are charged up to $$200$$ volts. A dielectric slab of thickness 4mm is inserted between the plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $$3.2mm$$. The dielectric constant of a dielectric slab is :

A number of identical condensers are first connected in parallel and then in series. The equivalent capacitance are found to be in the ratio $$9:1$$. The number of condensers used is :