## Wednesday, May 02, 2007

### Two Kerala Engineering/Medical Entrance 2007 Questions from Electrostatics

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper:

The plates of a parallel plate capacitor with air as medium are separated by a distance of 8 mm. A medium of dielectric constant 2 and thickness 4 mm having the same area is introduced between the plates. For the capacitance to remain the same, the distance between the plates is

(a) 8 mm (b) 6 mm (c) 4 mm (d) 12 mm (e) 10 mm

If a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space become respectively q/ε0A and q/Kε0A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε0A)(d–t)+(q/Kε0A)t = (q/ε0A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε0A/[d–t+(t/K)] = ε0A/[d–(t – t/K)].

Since the capacitance of the capacitor with air filling the entire space between the plates is ε0A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).

Therefore, the separation between the plates is to be increased by t–(t/K) = 4 – (4/2) = 2 mm.

The distance between the plates should be 8 + 2 = 10 mm.

The following question appeared in Kerala Medical Entrance 2007 question paper:

The capacitance of a parallel plate capacitor with air as medium is 3 μF. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 μF. The permittivity of the medium is

(a) 5 (b) 15 (c) 0.44×10–10C2N–1m–2

(d) 8.854×10–10C2N–1m–2 (e) 5C2N–1m–2

The capacitance of a parallel plate capacitor with air as the medium between the plates is ε0A/d and the capacitance with a medium of dielectric constant K instead of air is Kε0A/d where ε0, A and d are the relative permittivity of free space (or air very nearly), the area of each plate and the separation between the plates respectively. The capacitance therefore increases to K times the original value on replacing air with the dielectric medium.

Since the increase here is from 3 μF to 15 μF, the value of K is 5.

The permittivity of the medium is ε0K = 8.85×10–12×5 = 0.44×10–10C2N–1m–2.

[The value of ε0 was not given in the question paper. Some of you may not remember the value 8.854×10–12C2N–1m–2, but all of you are expected to remember the value of 1/4πε0, which is 9×109, very nearly. This will be enough to get the value of ε0].