JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 9)
Two capacitors, each having capacitance $$40 \,\mu \mathrm{F}$$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $$\mathrm{K}$$ such that the equivalence capacitance of the system became $$24 \,\mu \mathrm{F}$$. The value of $$\mathrm{K}$$ will be :
1.5
2.5
1.2
3
Explanation
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$${{40K \times 40} \over {40K + 40}} = 24$$
$$40K = 24(K + 1)$$
$$40K = 24K + 24$$
$$16K = 24$$
$$K = {{24} \over {16}} = {3 \over 2} = 1.5$$
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