Q. No.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. \(2.20\)
| 1. | increase. | 2. | decrease. |
| 3. | remain the same. | 4. | become zero. |
| 1. | \(KC \over 2(K+1)\) | 2. | \(2KC \over K+1\) |
| 3. | \(5KC \over 4K+1\) | 4. | \(4KC \over 3K+1\) |
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. |  |
A parallel plate capacitor with cross-sectional area \(A\) and separation \(d\) has air between the plates. An insulating slab of the same area but the thickness of \(\frac{d}{2}\) is inserted between the plates as shown in the figure having a dielectric constant, \(K=4\). The ratio of new capacitance to its original capacitance will be:
| 1. | \(2:1\) | 2. | \(8:5\) |
| 3. | \(6:5\) | 4. | \(4:1\) |
A parallel plate capacitor has capacitance \(C\). If it is equally filled with parallel layers of materials of dielectric constants \(K_1\) and \(K_2\), its capacity becomes \(C_1\). The ratio of \(C_1\) to \(C\) is:
| 1. | \(K_1 + K_2\) | 2. | \(\frac{K_{1} K_{2}}{K_{1}-K_{2}}\) |
| 3. | \(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\) | 4. | \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\) |
The dielectric constant of pure water is \(81\). Its permittivity will be: (in MKS units)
1. \(1.02\times 10^{-13}\)
2. \(8.86\times 10^{-12}\)
3. \(7.17\times 10^{-10}\)
4. \(7.8\times 10^{-10}\)
The insulation property of air breaks down at \(E = 3\times 10^{6}~\text{V/m}\). The maximum charge that can be given to a sphere of diameter \(5\) m is approximately:
1. \(2\times 10^{-5}~\text{C}\)
2. \(2\times 10^{-4}~\text{C}\)
3. \(2\times 10^{-3}~\text{C}\)
4. \(3\times 10^{-3}~\text{C}\)
A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness \(d_1\) and dielectric constant \(K_1\) and the other has thickness \(d_2\) and dielectric constant \(K_2\), as shown in the figure. This arrangement can be thought of as a dielectric slab of thickness \(d= d_1+d_2\) and effective dielectric constant \(K\). The \(K\) is:
| 1. | \(\frac{{K}_{1} {d}_{1}+{K}_{2} {d}_{2}}{{d}_{1}+{d}_{1}}\) | 2. | \(\frac{{K}_{1} {d}_{1}+{K}_{2} {d}_{2}}{{K}_{1}+{K}_{2}}\) |
| 3. | \(\frac{{K}_{1} {K}_{2}\left({d}_{1}+{d}_{2}\right)}{{K}_{1} {d}_{2}+{K}_{2} {d}_{1}}\) | 4. | \(\frac{2 {K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\) |
A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is \(A\) metre2 and the separation is \(t\) metre. The dielectric constants are \(k_1\) and \(k_2\) respectively. Its capacitance in farad will be:
1. \(\frac{\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
2. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} + k_{2}\right)}{2}\)
3. \(\frac{2\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
4. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} - k_{2}\right)}{2}\)
© 2024 GoodEd Technologies Pvt. Ltd.