Advertisements
Advertisements
Question
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?
Advertisements
Solution
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, `in_"r"` = 3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F
Distance between the plates is given by,
`"d" = "V"/"E"`
= `1000/10^6 = 10^-3 "m"`
Capacitance is given by relation
C = `(in_0in_"r""A")/"d"`
Where,
A = Area of each plate
`in_0` = Permittivity of free space = `8.85 xx 10^-12 "N"^-1 "C"^2 "m"^-2`
∴ `"A" = ("Cd")/(in_0in_"r")`
= `(50 xx 10^-12 xx 10^-3)/(8.85 xx 10^-12 xx 3) ≈ 19 "cm"^2`
Hence, the area of each plate is about 19 cm2.
APPEARS IN
RELATED QUESTIONS
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
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.
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.
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 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 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.
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?
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 _________.
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 is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
