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

Three capacitors each of capacity \(4\) µF are to be connected in such a way that the effective capacitance is \(6\) µF. This can be done by:

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)

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:

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 
 

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:

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. \(10~\mu\text{F}\)

Two parallel metal plates having charges +Q and –Q, face each other at a certain distance between them. If the plates are now dipped in the kerosene oil tank, the electric field between the plates will:

The electric potential V at any point (x, y, z), all in meters in space is given by V = $4{\mathrm{x}}^{\mathrm{2 }}$volt. The electric field at the point (1, 0, 2) in volt/meter, is:

Three charges, each \(+q\), are placed at the corners of an equilateral triangle \(ABC\) of sides \(BC\), \(AC\), and \(AB\). \(D\) and \(E\) are the mid-points of \(BC\) and \(CA\). The work done in taking a charge \(Q\) from \(D\) to \(E\) is: