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/ε_{0}A and q/Kε_{0}A 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/ε_{0}A)(d–t)+(q/Kε_{0}A)t = (q/ε_{0}A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is **C** = q/V = **ε _{0}A/[d–t+(t/K)]** =

**ε**

_{0}A/[d–(t – t/K)].Since the capacitance of the capacitor with air filling the entire space between the plates is ε_{0}A/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 ^{–10}**

**C**

^{2}N

^{–1}m^{–2}

**(d) 8.854****×10 ^{–10}**

**C**

^{2}N

^{–1}m^{–2}**(e) 5C**

^{2}N

^{–1}m^{–2}

The capacitance of a parallel plate capacitor with air as the medium between the plates is ε_{0}A/d and the capacitance with a medium of dielectric constant K instead of air is Kε_{0}A/d where ε_{0}, 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 ε_{0}K = 8.85×10^{–12}×5 = 0.44×10^{–10}C^{2}N^{–1}m^{–2}.

[The value of ε_{0} was not given in the question paper. Some of you may not remember the value 8.854×10^{–12}C^{2}N^{–1}m^{–2}, but all of you are expected to remember the value of 1/4πε_{0}, which is 9×10^{9}, very nearly. This will be enough to get the value of ε_{0}].

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