Advertisements
Advertisements
Question
An air-filled parallel-plate capacitor is to be constructed which can store 12 µC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is `3 xx 10^6 "Vm"^-1`
Advertisements
Solution
Charge that the capacitor can hold = 12 μC
Operating voltage = 1200 V
Breakdown strength, b = `3 xx 10^6 "Vm"^-1`
The separation between the plates of the capacitor is given by `d = V/b = 1200/(3 xx 10^6) = 4 xx 10^-4 "m"`
The capacitance of the capacitor is given by `C = Q/V = (12 xx 10^-6)/1200 = 10^-8 "F"`
The capacitance can be expressed in terms of the area of the plates (A) and the separation of the plates (d) as : `C = (∈_0A)/d = 10^-8 "F"`
Thus, the area of the plates is given by
`A = (10^-8 xx (4 xx 10^-4))/(8.854 xx 10^-12) = 0.45 "m"^2`
APPEARS IN
RELATED QUESTIONS
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports. Show that the capacitance of a spherical capacitor is given by
C = `(4piin_0"r"_1"r"_2)/("r"_1 - "r"_2)`
where r1 and r2 are the radii of outer and inner spheres, respectively.
Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are
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?
A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm and of radii 2 mm and 4 mm. (a) Calculate the capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii 4 mm and 8 mm. Calculate the capacitance.
Find the equivalent capacitances of the combinations shown in figure between the indicated points.
A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge of the capacitor to become 12.6 μC. Find the resistance of the circuit.
Each of the plates shown in figure has surface area `(96/∈_0) xx 10^-12` Fm on one side and the separation between the consecutive plates is 4⋅0 mm. The emf of the battery connected is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.
A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?
Figure shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l1 and l2. The left half of the dielectric slab has a dielectric constant K1 and the right half K2. Neglecting any friction, find the ration of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.
You are provided with 8 μF capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 μF.
Define ‘capacitance’. Give its unit.
Derive the expression for resultant capacitance, when the capacitor is connected in parallel.
The capacitance of a parallel plate capacitor is 60 µF. If the distance between the plates is tripled and area doubled then new capacitance will be ______.
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.
Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge on each capacitor.
Calculate equivalent capacitance of the circuit shown in the Figure given below:
The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness `3/8 d and d/2`, respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25 K, the value of K1 is:
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then:
- the charge stored in it. increases.
- the energy stored in it, decreases.
- its capacitance increases.
- the ratio of charge to its potential remains the same.
- the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
