The equivalent capacitance of the combination shown in the figure is:
1. | \(\dfrac{C}{2}\) | 2. | \(\dfrac{3C}{2}\) |
3. | \(3C\) | 4. | \(2C\) |
1. | \(\sqrt{\dfrac{R_1}{R_2}}\) | 2. | \(\dfrac{R^2_1}{R^2_2}\) |
3. | \(\dfrac{R_1}{R_2}\) | 4. | \(\dfrac{R_2}{R_1}\) |
Twenty seven drops of same size are charged at \(220~\text{V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop:
1. \(1520~\text{V}\)
2. \(1980~\text{V}\)
3. \(660~\text{V}\)
4. \(1320~\text{V}\)
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 new capacitance to its original capacitance will be:
1. | \(2:1\) | 2. | \(8:5\) |
3. | \(6:5\) | 4. | \(4:1\) |
1. | 2. | ||
3. | 4. |
1. | dependent on the material property of the sphere |
2. | more on the bigger sphere |
3. | more on the smaller sphere |
4. | equal on both the spheres |
1. | \(1.5\times 10^{-6}~\text{J}\) | 2. | \(4.5\times 10^{-6}~\text{J}\) |
3. | \(3.25\times 10^{-6}~\text{J}\) | 4. | \(2.25\times 10^{-6}~\text{J}\) |
A hollow metal sphere of radius \(R\) is given \(+Q\) charges to its outer surface. The electric potential at a distance \(\dfrac{R}{3}\) from the centre of the sphere will be:
1. \(\dfrac{1}{4\pi \varepsilon_0}\dfrac{Q}{9R}\)
2. \(\dfrac{3}{4\pi \varepsilon_0}\dfrac{Q}{R}\)
3. \(\dfrac{1}{4\pi \varepsilon_0}\dfrac{Q}{3R}\)
4. \(\dfrac{1}{4\pi \varepsilon_0}\dfrac{Q}{R}\)