Q 12-02-128JEE MainJEE Main 2022 (28 Jul, Shift 1)Easy
Two capacitors, each having capacitance $40\ \mu$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 $24\ \mu$F. The value of $K$ will be :
Answer: (A) $1.5$
$$\frac{1}{24} = \frac{1}{40K} + \frac{1}{40} \Rightarrow \frac{1}{40K} = \frac{5 - 3}{120} = \frac{1}{60}$$
$$40K = 60 \Rightarrow K = 1.5$$
Solution by Sreeraj P, M.Sc Physics