Electrostatic P...

Seven capacitors each of capacitance $$2 \;\mu $$F are to be connected in a configuration to obtain an effective capacitance of $$\frac{10}{11}\mu F$$. which of the combination shown in figure will achieve the desired result

Condensers of capacities 2$$\mu $$F and 3$$\mu $$F are connected in series and a condenser of capacity 1$$\mu $$F is connected in parallel with them. The resultant capacity is :

A potential difference of $$300$$ volts is applied to a combination of $$2.0\mu $$F and $$8.0\mu $$F capacitors connected in series. The charge on the $$2.0\mu $$F capacitor is :

5 condensers are connected as shown in the figure. The effective capacitance between ‘A’ and ‘B’ is :

The distance between the plates of a condenser is reduced to $$\frac{1}{4}th$$ and the space between the plates is filled up by a medium of dielectric constant K(2.8). The capacity is increased by :

The capacity of a parallel plate condenser with air medium is C . If half of the space between the plates is filled with a slab of dielectric constant K as shown in the figure, then the capacity becomes :

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

The effective capacitance between ‘A’ and ‘B’ is :

If a proton and an electron are accelerated through the same potential difference then:

Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. $$Q_{1}$$ and $$Q_{2}$$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $$q_{1}$$ and $$q_{2}$$ . Then :