The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become

1. $\frac{KC}{2\left(K+1\right)}$

2. $\frac{2KC}{K+1}$

3. $\frac{5KC}{4K+1}$

4. $\frac{4KC}{3K+1}$

An air capacitor of capacity $C=10\mu F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is

1. 120 $\mu C$

2. 699 $\mu C$

3. 480 $\mu C$

4. 24 $\mu C$

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

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