Question: Q. 2. A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1 and dielectric constant κ1 and the other has thickness d2 and dielectric constant κ2 as shown in Figure. This arrangement can be thought as a dielectric slab of thickness d(=d1+d2) and effective dielectric constant κ. The κ is :

(a) k1d1+k2d2d1+d2 (b) k1d1+k2d2k1+k2 (c) k1k2(d1+d2)(k1d1+k2d2) (d) 2k1k2k1+k2

[NCERT Exemplar]

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Solution:

Ans. Correct option : (c)

Explanation: Capacitance of a parallel plate capacitor filled with dielectric of constant k1 and thickness d1 is,

C1=k1ε0Ad1

Similarly, for other capacitance of a parallel plate capacitor filled with dielectric of constant k2 and thickness d2 is,

C2=k2ε0Ad2

Both capacitors are in series so equivalent capacitance C is related as :

(2) $C_{1}+\frac{1}{C_{2}}=\frac{d_{1}}{k_{1} \varepsilon_{0} A}+\frac{d_{2}}{\mathrm{k}{2} \varepsilon{0} A}$

(i)=1ε0A[k2d1+k1d2k1k2]

So, C=k1k2ε0A(k1d2+k2d1)

(ii)C=kε0Ad

where, d=(d1+d2)

So, multiply the numerator and denominator of eqn. (i) with (d1+d2),

(iii)C=k1k2ε0A(k1d2+k2d1)(d1+d2)(d1+d2)=k1k2(k1d2+k2d1)ε0A(d1+d2).

Comparing eqns. (ii) and (iii), the dielectric constant of new capacitor is :

k=k1k2(d1+d2)(k1d2+k2d1)



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