A parallel plate condenser has a uniform electric field \(E\)(V/m) in the space between the plates. If the distance between the plates is \(d\)(m) and area of each plate is \(A(\text{m}^2)\), the energy (joule) stored in the condenser is:
1. | \(\dfrac{1}{2}\varepsilon_0 E^2\) | 2. | \(\varepsilon_0 EAd\) |
3. | \(\dfrac{1}{2}\varepsilon_0 E^2Ad\) | 4. | \(\dfrac{E^2Ad}{\varepsilon_0}\) |
Two identical capacitors \(C_{1}\) and \(C_{2}\) of equal capacitance are connected as shown in the circuit. Terminals \(a\) and \(b\) of the key \(k\) are connected to charge capacitor \(C_{1}\) using a battery of emf \(V\) volt. Now disconnecting \(a\) and \(b\) terminals, terminals \(b\) and \(c\) are connected. Due to this, what will be the percentage loss of energy?
1. | \(75\%\) | 2. | \(0\%\) |
3. | \(50\%\) | 4. | \(25\%\) |
In a certain region of space with volume \(0.2\) m3, the electric potential is found to be \(5\) V throughout. The magnitude of electric field in this region is:
1. \(0.5\) N/C
2. \(1\) N/C
3. \(5\) N/C
4. zero
A short electric dipole has a dipole moment of \(16 \times 10^{-9} ~\text{C-}\text{m}\). The electric potential due to the dipole at a point at a distance of \(0.6~\text{m}\) from the centre of the dipole situated on a line making an angle of \(60^{\circ}\) with the dipole axis is: \(\left( \dfrac{1}{4\pi \varepsilon_0}= 9\times 10^{9}~\text{N-m}^2/\text{C}^2\right)\)
1. \(200~\text{V}\)
2. \(400~\text{V}\)
3. zero
4. \(50~\text{V}\)
The capacitance of a parallel plate capacitor with air as a medium is \(6~\mu\text{F}\). With the introduction of a dielectric medium, the capacitance becomes \(30~\mu\text{F}\). The permittivity of the medium is:\(\left(\varepsilon_0=8.85 \times 10^{-12} ~\text{C}^2 \text{N}^{-1} \text{m}^{-2}\right )\)
1. | \(1.77 \times 10^{-12}~ \text{C}^2 \text{N}^{-1} \text{m}^{-2}\) | 2. | \(0.44 \times 10^{-10} ~\text{C}^2 \text{N}^{-1} \text{m}^{-2}\) |
3. | \(5.00 ~\text{C}^2 \text{N}^{-1} \text{m}^{-2}\) | 4. | \(0.44 \times 10^{-13} ~\text{C}^2 \text{N}^{-1} \text{m}^{-2}\) |
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 \(\dfrac{d}{2}\) 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. | \(4:1\) |
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. | \(\dfrac{R_2}{R_1}\) |