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/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 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
RELATED QUESTIONS
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.]
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`.
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 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 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?
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.
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.
