Advertisements
Advertisements
प्रश्न
The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Advertisements
उत्तर
Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 × 10−4 m2
Distance between the plates, d = 2.5 mm = 2.5 × 10−3 m
Potential difference across the plates, V = 400 V
(a) Capacitance of the capacitor is given by the relation,
C = `(in_0"A")/"d"`
Electrostatic energy stored in the capacitor is given by the relation, `"E"_1 = 1/2 "CV"^2`
= `1/2 (in_0"A")/"d""V"^2`
Where,
`in_0` = Permittivity of free space = 8.85 × 10−12 C2 N−1 m−2
∴ `"E"_1 = (1 xx 8.85 xx 10^-12 xx 90 xx 10^-4 xx (400)^2)/(2 xx 2.5 xx 10^-3) = 2.55 xx 10^-6 "J"`
Hence, the electrostatic energy stored by the capacitor is `2.55 xx 10^-6 "J"`
(b) Volume of the given capacitor,
`"V'" = "A" xx "d"`
= `90 xx 10^-4 xx 25 xx 10^-3`
= `2.25 xx 10^-4 "m"^3`
Energy stored in the capacitor per unit volume is given by,
`"u" = "E"_1/("V'")`
= `(2.55 xx 10^-6)/(2.25 xx 10^-4) = 0.113 "J m"^-3`
Again, u = `"E"_1/("V'")`
= `((1/2)"CV"^2)/("Ad") = ((in_0"A")/(2"d")V^2)/("Ad") = 1/2in_0("V"/
"d")^2`
Where,
`"V"/"d"` = Electric intensity = E
∴ `"u"=1/2 in_0"E"^2`
APPEARS IN
संबंधित प्रश्न
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
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.
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.]
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.
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 has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:
1) the distance between the two plates is doubled?
2) a marble slab of dielectric constant 8 is introduced between the two plates?
Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
In a parallel plate capacitor, the capacity increases if ______.
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations:
- Key K is kept closed and plates of capacitors are moved apart using insulating handle.
- Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
- In A: Q remains same but C changes.
- In B: V remains same but C changes.
- In A: V remains same and hence Q changes.
- In B: Q remains same and hence V changes.
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.
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will ______.
