Electrostatic P...

If 3 capacitors of values 1, 2 and $$3 \mu F$$ are available. The maximum and minimum values of capacitance one can obtain by different combinations of the three capacitors together are respectively:

Three capacitors $$3 \mu F, 10 \mu F$$ and $$15 \mu F$$  are connected in series to a voltage source of 100V. The charge on $$15 \mu F$$ is :

A condenser is charged to a potential difference of  $$120 \; V$$. It's energy is $$1 \times 10^{-5}\; J$$. If battery is there and the space between plates is filled up with a dielectric medium $$\left ( \varepsilon  _{r}=5\right )$$. Its new energy is

A condenser of $$1\mu F$$ is charged to a potential of $$1000 V$$. If a dielectric slab of dielectric constant $$5$$ is introduced between the plates of the condenser after disconnecting the battery, the loss in the energy of the condenser is :

Two identical parallel plate capacitors are joined in series to $$100\;V$$ battery. Now a dielectric with $$K=4$$ is introduced between the plates of second capacitor. The potential difference on capacitors now becomes

The area of the positive plate is $$125\;cm^{2}$$ and the area of the negative plate is $$100\;cm^{2}$$, They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :

$$\left ( \varepsilon _{0}=8.9\times 10^{-12}\;C^{2}N^{-1}M^{-2} \right )$$

Two identical metal plates, separated by a distance d form a parallel plate capacitor. A metal sheet of thickness$$\dfrac{d}{2} $$ of the same area as that of either plate, is inserted between the plates. The ratio of the capacitance's after the insertion of the sheet to that before insertion is:

When a dielectric slab of thickness $$4 \;cm$$ is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by $$3 \; cm$$ to restore the capacity to it's original value. The dielectric constant of the slab is

Two capacitors of capacities 3$$\mu $$F and 6$$\mu F$$ are connected in series and connected to 120V. The potential differences across 3$$\mu $$F is $$V_{0}$$ and the charge here is $$q_{0}$$. We have :

A)$$q_{0}=40\mu C$$     B) $$V_{0}=60V$$

C)$$V_{0}=80V$$        D)$$q_{0}=240\mu C$$

Given four capacitors each of capacity 12$$\mu $$F. To get a capacity of 9$$\mu $$F, what combination can be used: