मराठी

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. - Physics

Advertisements
Advertisements

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? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.

Advertisements

उत्तर

The electric field between the plates is

`E=V/D`

The distance between plates is doubled, d' = 2d

`E'=(V')/(D')=(V/K)xx1/(2d)=1/2(E/K)`

Therefore, if the distance between the plates is double, the electric field will reduce to one half.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2014-2015 (March) Panchkula Set 3

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

The equivalent capacitance of the combination shown in the figure is _________ .


The separation between the plates of a parallel-plate capacitor is 0⋅500 cm and its plate area is 100 cm2. A 0⋅400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.


A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?


If the voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.


Calculate the resultant capacitances for each of the following combinations of capacitors.







Consider the following statements regarding series grouping of capacitors and select the correct statements.
  1. Charge on each capacitor remains same and equals to the main charge supplied by the battery.
  2. Potential difference and energy distribute in the reverse ratio of capacitance.
  3. Effective capacitance is even les than the least of teh individual capacitances.

Capacitors are used in electrical circuits where appliances need more ______.

A 5µF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 µF capacitor If the energy change during the charge redistribution is `"X"/100`J then value of X to the 100 nearest integer 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.
  1. Find the equivalent capacitance between points A and B in the given diagram.
  2. A dielectric slab is inserted between the plates of the parallel plate capacitor. The electric field between the plates decreases. Explain.
  3. 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.
     

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×