Q. No.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 \(C= 10~\mu\text{F}\) 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. \(\frac{Q^{2}}{2 \pi \varepsilon_{0} a}\)
| 1. | \(V={p\cos \theta \over 4 \pi \varepsilon_0r^2}\) | 2. | \(V={p\cos \theta \over 4 \pi \varepsilon_0r}\) |
| 3. | \(V={p\sin \theta \over 4 \pi \varepsilon_0r}\) | 4. | \(V={p\cos \theta \over 2 \pi \varepsilon_0r^2}\) |
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 \(K_1~\text{and}~K_2(K_{1} < K_{2})\) 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. |  |
| 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. \(\frac{C}{3}, \frac{V}{3}\)
2. \(3C, \frac{V}{3}\)
3. \(\frac{C}{3}, 3V\)
4. \(3C, 3V\)
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