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प्रश्न
Obtain the expression for capacitance for a parallel plate capacitor.
Derive an expression for the capacitance of a parallel plate capacitor with air present between the two plates.
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उत्तर १
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = `sigma/ε_0` where σ is the surface charge density on the plates σ = `"Q"/"A"`. If the separation distance d is very much smaller than the size of the plate (d2 << A), then the above result is used even for finite-sized
parallel plate capacitor.
Capacitance of a parallel plate capacitor
Capacitance of a parallel plate capacitor
The electric field between the plates is
E = `"Q"/("A"ε_0)` ....(1)
Since the electric field is unifonn, the electric potential between the plates having separation d is given by
V = Ed = `"Qd"/("A"ε_0)` ....(2)
Therefore the capacitance of the capacitor is given by
C = `"Q"/"V" = "Q"/(("Qd"/("A"ε_0))) = (ε_0"A")/"d"` ....(3)
From equation (3), it is evident that capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates. This can be understood from the following.
- If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
- If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant.
उत्तर २
Derivation of the expression for the capacitance.
Let the two plates be kept parallel to each other separated by a distance d and the cross-sectional area of each plate is A. Electric field by a single thin plate E = `sigma/(2epsi_0)`
The total electric field between the plates E = `sigma/epsi_0` = `Q/(Aepsi_0)`
The potential difference between the plates V = Ed = `[Q/(Aε_0)]`d.
Capacitance C = `"Q"/"V" = (Aepsi_0)/d`
संबंधित प्रश्न
A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) Field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.
A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.
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.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?
A capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charges? If no, what other information is needed?
Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be
Find the equivalent capacitance of the infinite ladder shown in figure between the points A and B.
A finite ladder is constructed by connecting several sections of 2 µF, 4 µF capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C. What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between?
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.
Calculate the resultant capacitances for each of the following combinations of capacitors.
Capacitors P and Q have identical cross-sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant Er as shown in the figure. Calculate the capacitance of capacitors P and Q.
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 ______.
Two spherical conductors A and B of radii a and b(b > a) are placed concentrically in the air. B is given a charge +Q and A is earthed. The equivalent capacitance of the system is ______.
Between the plates of parallel plate condenser there is 1 mm thick medium shoot of dielectric constant 4. It is charged at 100 volt. The electric field in volt/meter between the plates of capacitor is ______.
A 5µF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 µF capacitor If the energy change during the charge redistribution is `"X"/100`J then value of X to the 100 nearest integer is ______.
A capacitor has charge 50 µC. When the gap between the plate is filled with glass wool, then 120 µC charge flows through the battery to capacitor. The dielectric constant of glass wool 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.
