JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Capacitance

Two capacitors, each having capacitance

40 μ F

are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became

24 μ F

. The value of

K

will be :

(A)1.5✓ Correct
(B)2.5
(C)1.2
(D)3

Answer: (A)

Solution

One capacitor has capacitance

C = 40

μF, the other has

K C = 40 K

μF. For series combination:

\frac{1}{C _{e q}} = \frac{1}{C} + \frac{1}{K C}

, so

C_{e q} = \frac{K C}{K + 1}

. Given

C_{e q} = 24

μF:

24 = \frac{40 K}{K + 1}

. Solving:

24 (K + 1) = 40 K

24 K + 24 = 40 K

16 K = 24

K = 1.5

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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