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 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.
Advertisements
Solution
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/2 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)/2 + (E_0d)/2 =d/2 (E +E_0) (therefore t =d/2)`
`=> V=d/2 ((E_0)/K +E_0) (dE_0)/(2K) (K+1) (As,(E_0)/E = K)`
Now, `E_0 =σ/epsi_0 = q/(epsiA )=>V = d/(2K) q/(epsi_0A) (K +1)`
we know, `C = q/V = (2Kepsi_0A)/(d(K+1))`
APPEARS IN
RELATED QUESTIONS
Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery
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.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].
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.
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.
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.
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.
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.
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.
