Advertisements
Advertisements
Question
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.
Advertisements
Solution
Initially when there is vacuum between the two plates, the capacitance of the two parallel plates 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 = (2d)/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 = (2Ed)/3 +E_0 (d-(2d)/3) = (2Ed)/3 +(E_0d)/3 = d/3 (2E + E_0) (because t = d/2)`
`or , V = d/3 ((2E_0)/K + E_0) = (dE_0)/(3K) (K+2) (As,E_0/E = K)`
`Now , E_0 = σ/epsi_0 = q/(epsi_0A) => V = d/(3K) q/(epsi_0A) (K+2)`
`therefore C = q/V = (3K_(epsi_0 A))/(d(K+2))`.
APPEARS IN
RELATED QUESTIONS
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.
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/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.
For a one dimensional electric field, the correct relation of E and potential V is _________.
In a parallel plate capacitor, the capacity increases if ______.
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 filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will ______.
