CBSE Class 12 · 2020 · CapacitanceShort Answer

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The space between the plates of a parallel plate capacitor is completely filled in two ways. In the first case, it is filled with a slab of dielectric constant

K

. In the second case, it is filled with two slabs of equal thickness and dielectric constants

K_{1}

and

K_{2}

respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between

K

K_{1}

and

K_{2}

Answer: $K = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}}$ .

Solution

Case 1 — a single slab of constant

K

fills the whole gap

d

C_{1} = \frac{K ε _{0} A}{d}

Case 2 — two slabs, each of thickness

\frac{d}{2}

, act as two capacitors in series:

C_{a} = \frac{K _{1} ε _{0} A}{d /2} = \frac{2 K _{1} ε _{0} A}{d}, C_{b} = \frac{2 K _{2} ε _{0} A}{d}

In series,

\frac{1}{C _{2}} = \frac{1}{C _{a}} + \frac{1}{C _{b}} = \frac{d}{2 ε _{0} A} \frac{K _{1} + K _{2}}{K _{1} K _{2}}

C_{2} = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}} \frac{ε _{0} A}{d}

The two capacitances are equal,

C_{1} = C_{2}

\frac{K ε _{0} A}{d} = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}} \frac{ε _{0} A}{d}

K = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}}

CBSE Class 12 Board · 2020 · Subjective

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CBSE Class 12 · 2020 · CapacitanceShort Answer

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