A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in -

1. Reduction of charge on the plates and increase of potential difference across the plates

2. Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

3. Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

4. None of the above

The capacity of a condenser is 4 × 10^–6 farad and its potential is 100 volts. The energy released on discharging it fully will be -

1. 0.02 Joule

2. 0.04 Joule

3. 0.025 Joule

4. 0.05 Joule

The insulated spheres of radii R₁ and R₂ having charges Q₁ and Q₂ respectively are connected to each other. There is -

1. No change in the energy of the system

2. An increase in the energy of the system

3. Always a decrease in the energy of the system

4. A decrease in the energy of the system unless Q₁R₂ = Q₂R₁

The energy stored in a condenser of capacity C which has been raised to a potential V is given by -

1. $\frac{1}{2}CV$

2. $\frac{1}{2}C{V}^2$

3. CV

4. $\frac{1}{2VC}$

If two conducting spheres are separately charged and then brought in contact -

1. The total energy of the two spheres is conserved

2. The total charge on the two spheres is conserved

3. Both the total energy and charge are conserved

4. The final potential is always the mean of the original potentials of the two spheres

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is 

1. 8 times

2. 4 times

3. 2 times

4. 32 times

A parallel plate condenser has a capacitance 50 μF in air and 110 μF when immersed in an oil. The dielectric constant ‘k’ of the oil is 

1. 0.45

2. 0.55

3. 1.10

4. 2.20

Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes -

1. $\frac{{\epsilon }_{0}A}{d+t\left(1-\frac{1}{k}\right)}$

2. $\frac{{\epsilon }_{0}A}{d+t\left(1+\frac{1}{k}\right)}$

3. $\frac{{\epsilon }_{0}A}{d-t\left(1-\frac{1}{k}\right)}$

4. $\frac{{\epsilon }_{0}A}{d-t\left(1+\frac{1}{k}\right)}$

The capacity of parallel plate condenser depends on 

1. The type of metal used

2. The thickness of plates

3. The potential applied across the plates

4. The separation between the plates

The capacity of a parallel plate condenser is C. It's capacity when the separation between the plates is halved will be?

1. 4 C

2. 2 C

3. $\frac{C}{2}$

4. $\frac{C}{4}$