हिंदी

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.

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (March) Panchkula Set 3

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates . The capacitance now becomes _________ .


The plates of a parallel-plate capacitor are made of circular discs of radii 5⋅0 cm each. If the separation between the plates is 1⋅0 mm, what is the capacitance?


Each capacitor shown in figure has a capacitance of 5⋅0 µF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed?


A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.


Consider an assembly of three conducting concentric spherical shell of radii a, b and c as shown in figure Find the capacitance of the assembly between the points Aand B.


A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system ______.


A capacitor of 4 µ F is connected as shown in the circuit (Figure). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be ______.


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.


A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?


A parallel plate capacitor (A) of capacitance C is charged by a battery to voltage V. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C is connected across A. Find the ratio of final charges on A and B.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×