As per this diagram, a point charge \(\mathrm{+q}\) is placed at the origin \(\mathrm{O}.\) Work done in taking another point charge \(\mathrm{-Q}\) from the point \(\mathrm{A},\) coordinates \((\mathrm{0,a}),\) to another point \(\mathrm{B},\) coordinates \((\mathrm{a,0}),\) along the straight path \(\mathrm{AB}\) is:

      

1.	\( \left(\frac{-\mathrm{qQ}}{4 \pi \varepsilon_0} \frac{1}{\mathrm{a}^2}\right) \sqrt{2} \mathrm{a}\)	2.	zero
3.	\( \left(\frac{\mathrm{qQ}}{4 \pi \varepsilon_0} \frac{1}{\mathrm{a}^2}\right) \frac{1}{\sqrt{2}} \)	4.	\( \left(\frac{\mathrm{qQ}}{4 \pi \varepsilon_0} \frac{1}{\mathrm{a}^2}\right) \sqrt{2} \mathrm{a}\)

A network of four capacitors of capacity equal to ${\mathrm{C}}_1=\mathrm{C},{\mathrm{ C}}_2=2\mathrm{C},{\mathrm{ C}}_3=3\mathrm{C}\text{, and }{\mathrm{C}}_4=4\mathrm{C}$ is conducted to a battery as shown in the figure. The ratio of the charges on ${\mathrm{C}}_2\text{ and }{\mathrm{C}}_4$ is:

1. $\frac{7}{4}$

2. $\frac{22}{3}$

3. $\frac{3}{22}$

4. $\frac{4}{7}$

The effective capacity of the network between terminals \(\mathrm{A}\) and \(\mathrm{B}\) is:

1. \(6~\mu\text{F}~\)
2. \(20~\mu\text{F} ~\)
3. \(3~\mu\text{F}~\)
4. \(10~\mu\text{F}\)

Eight equally charged tiny drops are combined to form a big drop. If the potential on each drop is 10 V, then the potential of the big drop will be:

A bullet of mass 2 g is having a charge of 2 µC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m/s?
1. 50 kV
2. 5 V
3. 50 V
4. 5 kV 
 

An electric dipole has the magnitude of its charges as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively:

The energy and capacity of a charged parallel plate capacitor are \(E\) and \(C\) respectively. If a dielectric slab of ${}_{}$\(E_r=6\) is inserted in it, then the energy and capacity become:
(Assuming the charge on plates remains constant)

Some charge is being given to a conductor. Then it's potential:

A capacitor of capacity C₁ is charged up to V volt and then connected to an uncharged capacitor C₂. Then final P.D. across each will be:

1. $\frac{{\mathrm{C}}_2\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

2. $\frac{{\mathrm{C}}_1\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

3. $\left(1+\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}\right)$

4. $(1-\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1})\mathrm{V}$

If identical charges (–q) are placed at each corner of a cube of side 'b' then the electrical potential energy of charge (+q) which is placed at centre of the cube will be:

1. $\frac{-4\sqrt{2}{\mathrm{q}}^2}{{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

2. $\frac{-8\sqrt{2}{\mathrm{q}}^2}{{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

3. $\frac{-4{\mathrm{q}}^2}{\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{b}}$

4. $\frac{8\sqrt{2}{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{b}}$