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:
1. | \(KC \over 2(K+1)\) | 2. | \(2KC \over K+1\) |
3. | \(5KC \over 4K+1\) | 4. | \(4KC \over 3K+1\) |
An air capacitor of capacity 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
2. 699
3. 480
4. 24
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.
3.
4.
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)
1. | \(V={pcos \theta \over 4 \pi \varepsilon_0r^2}\) | 2. | \(V={pcos \theta \over 4 \pi \varepsilon_0r}\) |
3. | \(V={psin \theta \over 4 \pi \varepsilon_0r}\) | 4. | \(V={pcos \theta \over 2 \pi \varepsilon_0r^2}\) |
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 |
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 () 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:
1. | Zero and \(\mathrm{Q} / 4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}^2\) |
2. | \(\mathrm{Q} / 4 \pi \varepsilon_{\mathrm{O}} \mathrm{R}\) and zero |
3. | \(\mathrm{Q} / 4 \pi \varepsilon_{\mathrm{O}} \mathrm{R}\) and \(\mathrm{Q} / 4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}^2\) |
4. | Both are zero |
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:
1. | \(Q=-q \) | 2. | \(Q=-\frac{1}{q} \) |
3. | \(Q=q \) | 4. | \(\mathrm{Q}=\frac{1}{q}\) |
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.
2.
3.
4.