Advertisements
Advertisements
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.
Advertisements
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
-
A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
-
Let A be the area of each plate and d the separation between them. The two plates have charges Q and −Q.
-
Surface charge density of plate 1, σ = Q/A, and that of plate 2 is σ.
-
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
-
The direction of electric field is from the positive to the negative plate.
-
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 `
APPEARS IN
संबंधित प्रश्न
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
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?
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.
Define the capacitance of a capacitor and its SI unit.
A parallel-plate capacitor is filled with a dielectric material of resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.
Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
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 _________.
