JEE PYQ: Capacitance Question 35
Question 35 - 2019 (10 Jan Shift 1)
A parallel plate capacitor is of area 6 cm$^2$ and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10$, $K_2 = 12$ and $K_3 = 14$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:
(1) 4
(2) 14
(3) 12
(4) 36
Show Answer
Answer: (3)
Solution
Let dielectric constant of material used be K. $\frac{K\epsilon_0 A}{d} = \frac{K_1 \epsilon_0 A_1}{d} + \frac{K_2 \epsilon_0 A_2}{d} + \frac{K_3 \epsilon_0 A_3}{d}$ (in series configuration). Using harmonic mean for series: $\frac{\epsilon_0 A}{d}\left(\frac{10}{3} + \frac{12}{3} + \frac{14}{3}\right)^{-1}$… Actually: $K = \frac{10 + 12 + 14}{3} = 12$ (for parallel arrangement of equal thickness slabs).
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