Electrostatic Potential and Capacitance: CapacitorsContact Number: 8527521718Electrostatic Potential and Capacitance: CapacitorsContact Number: 8527521718A 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. | \(4:1\) |
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. \(0.44 \times 10^{-13} ~\text{C}^2 \text{N}^{-1} \text{m}^{-2}\)
A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:
1. decreases.
2. does not change.
3. becomes zero.
4. increases.
A parallel plate air capacitor has capacitance \(C,\) the distance of separation between plates is \(d\) and potential difference \(V\) is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:
| 1. | \(\frac{C^2V^2}{2d}\) | 2. | \(\frac{CV^2}{2d}\) |
| 3. | \(\frac{CV^2}{d}\) | 4. | \(\frac{C^2V^2}{2d^2}\) |
The energy required to charge a parallel plate condenser of plate separation, \(d\) and plate area of cross-section, \(A\) such that the uniform electric field between the plates is \(E,\) is:
| 1. | \(\dfrac{\varepsilon_0E^2}{2Ad}\) | 2. | \(\dfrac{\varepsilon_0E^2}{Ad}\) |
| 3. | \(\varepsilon_0E^2Ad\) | 4. | \(\dfrac{1}{2}\varepsilon_0E^2Ad\) |
The capacity of a parallel plate condenser is C. It's capacity when the separation between the plates is halved will be?
1. 4 C
2. 2 C
3.
4.
The plates of a parallel plate condenser are pulled apart with a velocity v. If at any instant their mutual distance of separation is d, then the magnitude of the time rate of change of capacity depends on d as follows
(1) 1/d
(2) 1/d2
(3) d2
(4) d
The capacity of a parallel plate condenser is \(15~\mu\text{F}\), when the distance between its plates is \(6~\text{cm}\). If the distance between the plates is reduced to \(2~\text{cm}\), then the capacity of this parallel plate condenser will be?
1. \(15~\mu\text{F}\)
2. \(30~\mu\text{F}\)
3. \(45~\mu\text{F}\)
4. \(60~\mu\text{F}\)
The equivalent capacitance between \(A\) and \(B\) is:
| 1. | \(2~\mu\text{F}\) | 2. | \(3~\mu\text{F}\) |
| 3. | \(5~\mu\text{F}\) | 4. | \(0.5~\mu\text{F}\) |
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. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} - k_{2}\right)}{2}\)
Three capacitors of capacitances \(3~\mu\text{F}\), \(9~\mu\text{F}\) and \(18~\mu\text{F}\) are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases \(\frac{C_s}{C_p}\) will be:
1. \(1:15\)
2. \(15:1\)
3. \(1:1\)
4. \(1:3\)
A parallel plate capacitor of capacitance \(C\) is connected to a battery and is charged to a potential difference \(V\). Another capacitor of capacitance \(2C\) is connected to another battery and is charged to potential difference \(2V.\) The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is?
1. zero
2. \(\frac{25 C V^{2}}{6}\)
3. \(\frac{3 C V^{2}}{2}\)
4. \(\frac{9 C V^{2}}{2}\)
In the connections shown in the adjoining figure, the equivalent capacity between \(A\) and \(B\) will be:
1. \(10.8~\mu\text{F}\)
2. \(69~\mu\text{F}\)
3. \(15~\mu\text{F}\)
4. \(10~\mu\text{F}\)
Two capacitances of capacity C1 and C2 are connected in series and potential difference V is applied across it. Then the potential difference across C1 will be:
1.
2.
3.
4.
A series combination of \(n_1\) capacitors, each of value \(C_1\), is charged by a source of potential difference \(4\) V. When another parallel combination of \(n_2\) capacitors, each of value \(C_2\), is charged by a source of potential difference \(V\), it has the same (total) energy stored in it as the first combination has. The value of \(C_2\) in terms of \(C_1\) is:
1. \(\frac{2C_1}{n_1n_2}\)
2. \(16\frac{n_2}{n_1}C_1\)
3. \(2\frac{n_2}{n_1}C_1\)
4. \(\frac{16C_1}{n_1n_2}\)
Two condensers, one of capacity \(C\) and the other of capacity \(\frac{C}2\) are connected to a \(V\) volt battery, as shown in the figure.
The energy stored in the capacitors when both condensers are fully charged will be:
1. \(2CV^2\)
2. \({1 \over4}CV^2\)
3. \({3 \over4}CV^2\)
4. \({1 \over2}CV^2\)
| 1. | \(10^{7}\) joule and \(300\) paise |
| 2. | \(5\times 10^{6}\) joule and \(300\) paise |
| 3. | \(5\times 10^{6}\) joule and \(150\) paise |
| 4. | \(10^7\) joule and \(150\) paise |
The capacities of two conductors are C1 and C2 and their respective potentials are V1 and . If they are connected by a thin wire, then the loss of energy will be given by
1.
2.
3.
4.
A capacitor of capacitance 5 μF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω. The amount of charge on the capacitor plate is?
1. 0 μC
2. 5 μC
3. 10 μC
4. 25 μC
In the given figure each plate of capacitance C has partial value of charge equal to:
1. CE
2.
3.
4.
