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`
संबंधित प्रश्न
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
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.]
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1/2)` QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor `1/2`.
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 \[\sqrt{3}\].
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3m2 and the separation between the plates is 3 mm.
- Calculate the capacitance of the capacitor.
- If this capacitor is connected to 100 V supply, what would be the charge on each plate?
- How would charge on the plates be affected, if a 3 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?
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.
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 is filled with a dielectric material of resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.
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?
Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
For a one dimensional electric field, the correct relation of E and potential V is _________.
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 ______.
