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प्रश्न
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.
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उत्तर
The capacitance of a capacitor is the amount of charge which creates unit potential difference between collecting plate and condensing plate after giving charge on the collecting plate.
Parallel Plate Capacitor
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A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
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Let A be the area of each plate and d the separation between them. The two plates have charges Q and −Q.
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Surface charge density of plate 1, σ = Q/A, and that of plate 2 is σ.
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Electric field in different regions:
Outer region I,
`E = σ/(2ε_0) - σ/(2ε_0 ) = 0`
In the inner region between plates 1 and 2, the electric fields due to the two charged plates add up. So,
\[E = \frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}\]
E=σ2ε0+σ2ε0=σε0=Qε0
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The direction of electric field is from the positive to the negative plate.
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For uniform electric field, potential difference is simply the electric field multiplied by the distance between the plates, i.e.
`V=Ed = 1/c (Qd)/A`
Capacitance C of the parallel plate capacitor in vacuum is
`C = Q/V =(ε_0A)/d `
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संबंधित प्रश्न
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
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.
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`.
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.
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.
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:
What is the effect of increasing plate separation \[d\] on capacitance?
For the same voltage, what change allows more charge to be stored and more energy to be stored in the capacitor?
