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प्रश्न
A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (b) The energy stored in the capacitor Justify your answer by writing the necessary expressions
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उत्तर
As the capacitance of the capacitor,
`C'=(in_0KA)/(d')=(in_0KA)/(2d)=1/2C " ...1"`
Energy stored in the capacitor is
`U=Q^2/(2C)`
`U'=Q^2/(2C')=Q^2/(2(1/2)C)=2(Q^2/(2C))2U " from 1"`
Therefore, when the distance between the plates is doubled, the capacitance reduces to half. Therefore, energy stored in the capacitor becomes double.
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संबंधित प्रश्न
Following operations can be performed on a capacitor:
X − connect the capacitor to a battery of emf ε.
Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.
Find the charges on the three capacitors connected to a battery as shown in figure.
Take `C_1 = 2.0 uF , C_2 = 4.0 uF , C_3 = 6.0 uF and V` = 12 volts.
Each of the capacitors shown in figure has a capacitance of 2 µF. find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is ______.
The radius of a sphere of capacity 1 microfarad in the air is ______
Between the plates of parallel plate condenser there is 1 mm thick medium shoot of dielectric constant 4. It is charged at 100 volt. The electric field in volt/meter between the plates of capacitor is ______.
A capacitor of capacity 2 µF is charged to a potential difference of 12 V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit at a time when the potential difference across the capacitor is 6.0 V is ______ × 10-1A.
Read the following paragraph and answer the questions.
| A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain the desired value of capacitance. |
- Find the equivalent capacitance between points A and B in the given diagram.
- A dielectric slab is inserted between the plates of the parallel plate capacitor. The electric field between the plates decreases. Explain.
- A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
OR
Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in the figure. Find an expression for the capacitance of the system.
