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 :

Solution

Physics sir · 4 steps

Only one capacitor changes, so take the series formula, peel off the untouched 40 µF and see what is left.

1. Dielectric effect. Filling a capacitor completely with a dielectric multiplies its capacitance by K. So one capacitor becomes 40K μF and the other stays 40 μF.

   1/24 = 1/40K + 1/40

2. Isolate the unknown. 1/40K = 1/24 - 1/40 = (5 - 3)/120 = 1/60. So the filled capacitor is 40K = 60 μF.
3. Sense check. Without the dielectric, two 40 μF capacitors in series give 20 μF. A dielectric can only increase capacitance, so a rise to 24 μF makes sense.
4. So. K = 60/40 = 1.5, which is option (a).

The idea to keep. In series, add reciprocals. When only one capacitor is changed, subtract the known reciprocal from 1/C_(eq) to get that capacitor alone, then compare it with its old value to find K.

Answer from the NTA final answer key. Solution by PrepWiser.