A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is \(A\) metre² and the separation is \(t\) metre. The dielectric constants are \(k_1\) and \(k_2\) respectively. Its capacitance in farad will be:

        
1. \(\frac{\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
2. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} + k_{2}\right)}{2}\)
3. \(\frac{2\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
4. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} - k_{2}\right)}{2}\)

Three condensers each of capacitance 2F are put in series. The resultant capacitance is 

1. 6F

2. $\frac{3}{2}F$

3. $\frac{2}{3}F$

4. 5F

Four condensers are joined as shown in the adjoining figure. The capacity of each is 8 μF. The equivalent capacity between the points A and B will be

1. 32 μF

2. 2 μF

3. 8 μF

4. 16 μF

The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points A and B is 60 volts. Then the equivalent capacity between A and B and the charge on 5 μF capacitance will be respectively

1. 44 μF; 300 μC

2. 16 μF; 150 μC

3. 15 μF; 200 μC

4. 4 μF; 50 μC

Four plates of the same area of cross-section are joined as shown in the figure. The distance between each plate is d. The equivalent capacity across A and B will be

1. $\frac{2{\epsilon }_{0}A}{d}$

2. $\frac{3{\epsilon }_{0}A}{d}$

3. $\frac{3{\epsilon }_{0}A}{2d}$

4. $\frac{{\epsilon }_{0}A}{d}$

In the adjoining figure, four capacitors are shown with their respective capacities and the P.D. applied. The charge and the P.D. across the 4 μF capacitor will be

1. 600 μC; 150 volts

2. 300 μC; 75 volts

3. 800 μC; 200 volts

4. 580 μC; 145 volts

In the connections shown in the adjoining figure, the equivalent capacity between \(A\) and \(B\) will be:

1. \(10.8~\mu\text{F}\)
2. \(69~\mu\text{F}\)
3. \(15~\mu\text{F}\)
4. \(10~\mu\text{F}\)

2 μF capacitance has potential difference across its two terminals 200 volts. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then P.D. becomes 20 volts. Then the capacity of another capacitance will be 

1. 2 μF

2. 4 μF

3. 18 μF

4. 10 μF

The resultant capacitance across 300 v battery in the figure shown is equal to

1. 1 μF

2. $\frac{31}{120}$ μF

3. 2 μF

4. $\frac{120}{31}$μF

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system 

1. increases by a factor of 4

2.decreases by a factor of 2

3. remain the same 

4. increases by a factor of 2