Two concentric conducting spherical shells carry charge Q each. The inner shell is earthed. The charge that flows into the earth is:

   

1. Q

2. $\frac{3\mathrm{Q}}{2}$

3. $\frac{-\mathrm{Q}}{2}$

4. $\frac{-3\mathrm{Q}}{2}$

The equivalent capacitance between the points A and B in the given network is:

1. 25 $\mathrm{\mu F}$

2. 16 $\mathrm{\mu F}$

3. 21 $\mathrm{\mu F}$

4. 12 $\mathrm{\mu F}$

Five equal capacitors connected in series have a resultant capacitance of $\mathrm{4 }\mu F$. The total energy stored in these when these are connected in parallel and charged to 400 V is:
1. 1 J
2. 8 J
3. 16 J
4. 4 J

Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of V₀ volt. If one of the capacitors is completely immersed in a liquid with dielectric constant K, the potential difference between the plates of the other capacitor will change to:

1. \(\frac{K +   1}{K} V_{0}\)

2. \(\frac{K}{K +   1} V_{0}\)

3. \(\frac{K +   1}{2 K} V_{0}\)

4. \(\frac{2 K}{K +   1} V_{0}\)

A parallel plate air capacitor has a capacity of C, the distance of separation between plates is d and potential difference V is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:

1. $\frac{C^2{V}^2}{2d}$

2. $\frac{C{V}^2}{2d}$

3. $\frac{C{V}^2}{d}$

4. $\frac{C^2{V}^2}{2d^2}$

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

In the given circuit if point C is connected to the earth and a potential of +2000 V is given to the point A, the potential at B is:
            

The figure shows some of the equipotential surfaces. Magnitude and direction of the electric field is given by:

When a negative charge is released and moves in the electric field, it moves towards a position of:

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. $2.21\times {10}^9$