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
As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?
Each of the capacitors shown in figure has a capacitance of 2 µF. find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.
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.
A capacitor of capacitance 2⋅0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4⋅0 µF as shown in figure . Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.
Consider the situation shown in the figure. The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery.(c) Find the change in energy stored in the capacitors.(d) Find the heat developed in the system.
A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.
A sphercial capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K up to a radius c as shown in figure . Calculate the capacitance.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is ______.
For the given capacitor configuration
- Find the charges on each capacitor
- potential difference across them
- energy stored in each capacitor.
- Charge on each capacitor remains same and equals to the main charge supplied by the battery.
- Potential difference and energy distribute in the reverse ratio of capacitance.
- Effective capacitance is even les than the least of teh individual capacitances.
The work done in placing a charge of 8 × 10–18 coulomb on a condenser of capacity 100 micro-farad is ______.
A capacitor of 4 µ F is connected as shown in the circuit (Figure). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be ______.
Current versus time and voltage versus time graphs of a circuit element are shown in figure.
 |
 |
The type of the circuit element is ______.
Read the following paragraph and answer the questions.
| A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain the desired value of capacitance. |
- Find the equivalent capacitance between points A and B in the given diagram.
- A dielectric slab is inserted between the plates of the parallel plate capacitor. The electric field between the plates decreases. Explain.
- A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
OR
Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in the figure. Find an expression for the capacitance of the system.
Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge on each capacitor.
