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

       

1. \(25~\mu\text{F}\)
2. \(16~\mu\text{F}\)
3. \(21~\mu\text{F}\)
4. \(12~\mu\text{F}\)

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:

In the given circuit if point \(C\) is connected to the earth and a potential of \(+2000~\text{V}\) is given to the point \(A\), the potential at \(B\) is:
             

The figure shows some of the equipotential surfaces. The magnitude and direction of the electric field are 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 \(V = 4~\text{volt}\) each plate of the capacitor has a surface area of\(10^{-2}~\text{m}^2\) and the plates are$\mathrm{}$ \(0.1\times10^{-3}~\text{m}\)apart, then the number of excess electrons on the negative plate is:

An air capacitor of capacity \(C= 10~\mu\text{F}\) 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. \(120~\mu\text{C}\)
2. \(699~\mu\text{C}\)
3. \(480~\mu\text{C}\)
4. \(24~\mu\text{C}\)

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. \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\)
3. \(\frac{\sqrt{2} Q^{2}}{\pi \varepsilon_{0} a}\)
4. \(\frac{Q^{2}}{2 \pi \varepsilon_{0} a}\)