Four equal charges Q are placed at the four corners of a square of each side ‘a’. Work done in removing a charge – Q from its centre to infinity is:

1. 0

2. $\frac{\sqrt{2}{Q}^2}{4\pi {\epsilon }_{0}a}$

3. $\frac{\sqrt{2}{Q}^2}{\pi {\epsilon }_{0}a}$

4. $\frac{Q^2}{2\pi {\epsilon }_{0}a}$

Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by: (r >> 2a) (Where p = 2qa) 

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 \(50~\text{V}\)?

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 K₁&K₂ ($K_1<K_2$) 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.

A conducting sphere of radius R is given a charge Q. The electric potential and field at the center of the sphere respectively are:

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. $\frac{{\epsilon }_{0}AV}{d}, \frac{2{\epsilon }_{0}AV}{d}$

2. $\frac{{\epsilon }_{0}AV}{d}, \frac{2{\epsilon }_{0}AV}{d}$

3. $$ $\frac{{\epsilon }_{0}AV}{d}, \frac{-2{\epsilon }_{0}AV}{d}$

4. $\frac{-{\epsilon }_{0}AV}{d}, \frac{-2{\epsilon }_{0}AV}{d}$

A network of four capacitors of capacity equal to $C_1=C,$ $C_2=2C,$ $C_3=3C$ and C₄ = 4C are connected in a battery as shown in the figure. The ratio of the charges on C₂ and C₄ is:
       
1. $\frac{22}{3}$

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

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

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