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

The diagrams below show regions of equipotentials.

An electric dipole is place at an angle of \(30^{\circ}\) with an electric field intensity \(2\times10^{5}~\text{N/C}\). It experiences a torque equal to \(4~\text{Nm}\). The charge on the dipole, if the dipole length is \(2~\text{cm}\), is: 

A parallel-plate capacitor of area A, plate separation d, and capacitance C is filled with four dielectric materials having dielectric constants $k_1,k_2,k_3$ $$and $k_4$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

1. $\mathrm{k}={\mathrm{k}}_1+{\mathrm{k}}_2+{\mathrm{k}}_3+3{\mathrm{k}}_4$

2. $k=\frac{2}{3}\left(k_1+k_2+k_3\right)+2k_4$

3. $\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2k_4}$

4. $\frac{1}{\mathrm{k}}=\frac{1}{{\mathrm{k}}_1}+\frac{1}{{\mathrm{k}}_2}+\frac{1}{{\mathrm{k}}_3}+\frac{3}{2{\mathrm{k}}_4}$

A capacitor of \(2~\mu\text{F}\) is charged as shown in the figure. When the switch \(S\) is turned to position \(2\), the percentage of its stored energy dissipated is: