Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q₃ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $\frac{{\mathrm{q}}_3}{4\mathrm{\pi }{\in }_0}\mathrm{k}$ , where k is:

.

1. 8q₂

2. 6q₂

3. 8q₁

4. 6q₁

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:

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:

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

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 \(C_1\) is charged up to \(V\) volt and then connected to an uncharged capacitor \(C_2\). Then final P.D. across each will be:
1. \(\frac{C_{2} V}{C_{1} + C_{2}}\)
2. \(\frac{C_{1} V}{C_{1} + C_{2}}\)
3. \(\left(1 + \frac{C_{2}}{C_{1}}\right)\)
4. \(\left(1 - \frac{C_{2}}{C_{1}} \right) V\)