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. \(2.20\)
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
1. increase
2. decrease
3. remain the same
4. become zero
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
| 1. | \(KC \over 2(K+1)\) | 2. | \(2KC \over K+1\) |
| 3. | \(5KC \over 4K+1\) | 4. | \(4KC \over 3K+1\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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 the electric field \(E\) between the plates with distance \(d\) as measured from the plate \(P\) is correctly shown by:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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 \(\dfrac{d}{2}\) is inserted between the plates as shown in the figure, having a dielectric constant, \(K=4.\) The ratio of the new capacitance to its original capacitance will be:
| 1. | \(2:1\) | 2. | \(8:5\) |
| 3. | \(6:5\) | 4. | \(4:1\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. | \(\dfrac{{K}_{1} {d}_{1}+{K}_{2} {d}_{2}}{{d}_{1}+{d}_{1}}\) | 2. | \(\dfrac{{K}_{1} {d}_{1}+{K}_{2} {d}_{2}}{{K}_{1}+{K}_{2}}\) |
| 3. | \(\dfrac{{K}_{1} {K}_{2}\left({d}_{1}+{d}_{2}\right)}{{K}_{1} {d}_{2}+{K}_{2} {d}_{1}}\) | 4. | \(\dfrac{2 {K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. | \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. \(7.8\times 10^{-10}\)
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. \(3\times 10^{-3}~\text{C}\)
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} - k_{2}\right)}{2}\)
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
© 2026 GoodEd Technologies Pvt. Ltd.