Advertisements
Advertisements
प्रश्न
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/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.
Advertisements
उत्तर
Initially when there is vacuum between the two plates, the capacitance of the two parallel plate is, `C_0 = (epsi_0A)/d`where, A is the area of parallel plates.
Suppose, that the capacitor is connected to a battery, an electric field E0 is produced.
Now, if we insert the dielectric slab of thickness `t=d/3`, the electric field reduces to E.
Now the gap between plates is divided in two parts, for distance t there is electric field E and for the remaining distance (d– t) the electric field is E0.
If V be the potential difference between the plates of the capacitor, then V = Et + E0(d–t)
`V = (Ed)/3 + E_0 (d-d/3)=(Ed)/3 +E_0((2d)/3) = d/3 (E +2E_0) (therefore t=d/2)`
`or, V = d/3 (E_0/K +2E_0) (dE_0)/(3K)(2K +1) (As,E_0/E =K)`
Now,`E_0 = σ/epsi_0 = q/(epsi_0A) => V = d/(3K )q/(epsi_0A) (2K +1)`
We know, `C = q/V = (3Kepsi_0A)/(d(2K+1))`
APPEARS IN
संबंधित प्रश्न
Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.
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.]
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.
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\frac{3d}{4}\]. Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric.
A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor
A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:
1) the distance between the two plates is doubled?
2) a marble slab of dielectric constant 8 is introduced between the two plates?
In a parallel plate capacitor, the capacity increases if ______.
Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
