In the given figure if $\mathrm{V}=4 v\mathrm{olt}$, each plate of the capacitor has a surface area of ${10}^{-\mathrm{2 }}{\mathrm{m}}^2$ and the plates are $0.1\times {10}^{-\mathrm{3 }}\mathrm{m}$ apart, then the number of excess electrons on the negative plate is:

           

1. $5.15\times {10}^9$

2. $2.21\times {10}^{10}$

3. $3.33\times {10}^9$

4. $2.21\times {10}^9$

The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then the new capacitance will be: 

An air capacitor of capacity $C=\mathrm{10 }\mu F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is:

1. 120 $\mu C$

2. 699 $\mu C$

3. 480 $\mu C$

4. 24 $\mu C$

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: