Advertisements
Advertisements
प्रश्न
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Advertisements
उत्तर
Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = 0.5 × 10−2 m
Capacitance of a parallel plate capacitor is given by the relation,
C = `(in_0"A")/"d"`
A = `("Cd")/in_0`
Where,
`in_0` = Permittivity of free space = 8.85 × 10−12 C2 N−1 m−2
∴ A = `(2 xx 0.5 xx 10^-2)/(8.85 xx 10^-12)` = 1130 km2
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.
APPEARS IN
संबंधित प्रश्न
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\frac{3d}{4}\]. Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric.
A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor
Define the capacitance of a capacitor and its SI unit.
A parallel-plate capacitor with plate area 20 cm2 and plate separation 1.0 mm is connected to a battery. The resistance of the circuit is 10 kΩ. Find the time constant of the circuit.
A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
For a one dimensional electric field, the correct relation of E and potential V is _________.
Two charges – q each are separated by distance 2d. A third charge + q is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.
What is a capacitor that consists of two large, parallel, conducting plates separated by a small distance called?
What is the surface charge density \[\sigma\] on a plate carrying charge \[Q\] over area \[A\]?
What is the effect of increasing plate area \[A\] on capacitance, and why?
When a dielectric slab of dielectric constant \[K\] is inserted between the plates, what is the electric field \[E'\] due to polarisation of the dielectric?
For the same voltage, what change allows more charge to be stored and more energy to be stored in the capacitor?
How are capacitors used for energy storage in power supplies?
