How much kinetic energy will be gained by an \(\alpha\text-\text{particle}\) in going from a point at \(70~\text{V}\) to another point at \(50~\text{V}\)?

1.	\(40~\text{eV}\)	2.	\(40~\text{keV}\)
3.	\(40~\text{MeV}\)	4.	0

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A parallel plate condenser has a capacitance \(50~\mu\text{F}\) in air and \(110~\mu\text{F}\) when immersed in an oil. The dielectric constant \(k\) of the oil is: 
1. \(0.45\)
2. \(0.55\)
3. \(1.10\)
4. \(2.20\)

Two thin dielectric slabs of dielectric constants \(K_1~\text{and}~K_2(K_{1} < K_{2})\) are inserted between plates of a parallel capacitor, as shown in the figure. The variation of electric field \(E\) between the plates with distance \(d\) as measured from plate \(P\) is correctly shown by:  
  

1.		2.
3.		4.

Four point charges \(-Q, -q,2q~\text{and}~2Q\)$$ are placed, one at each corner of the square. The relation between \(Q\) and \(q\) for which the potential at the center of the square is zero, is:

Three capacitors each of capacitance \(C\) and of breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be:
1. \(\frac{C}{3}, \frac{V}{3}\)
2. \(3C, \frac{V}{3}\)
3. \(\frac{C}{3}, 3V\)
4. \(3C, 3V\)

Five identical plates each of area \(A\) are joined as shown in the figure. The distance between the plates is \(d\). The plates are connected to a potential difference of \(V\) volts. The charge on plates \(1\) and \(4\) will be:

1. \(-\frac{\varepsilon_{0} A V}{d} ,  \frac{2\varepsilon_{0} A V}{d}\)
2. \(\frac{\varepsilon_{0} A V}{d} ,  \frac{2\varepsilon_{0} A V}{d}\)
3. \(\frac{\varepsilon_{0} A V}{d} , -\frac{2\varepsilon_{0} A V}{d}\)
4. \(-\frac{\varepsilon_{0} A V}{d} ,  -\frac{2\varepsilon_{0} A V}{d}\)

A network of four capacitors of capacity equal to \(C_1 = C, C_2 = 2C, C_3 = 3C\) and \(C_4 = 3C\) are connected in a battery as shown in the figure. The ratio of the charges on \(C_2\) and \(C_4\) is:
       
1. \(\frac{22}{3}\)
2. \(\frac{3}{22}\)
3. \(\frac{7}{4}\)
4. \(\frac{4}{7}\)

An electric dipole of moment \(p\) is placed in an electric field of intensity \(E\). The dipole acquires a position such that the axis of the dipole makes an angle \(\theta\) with the direction of the field. Assuming that the potential energy of the dipole to be zero when \(\theta = 90^{\circ},\) the torque and the potential energy of the dipole will respectively be:

The equivalent capacitance between points \(a\) and \(b\) in the network shown below is:

1. \(5~\text{C}\)
2. \(4~\text{C}\)
3. \(3~\text{C}\)
4. \(2~\text{C}\)