Two parallel plate capacitors, each of the capacitance 40 μF are connected in series. The space between the plates of one capacitor is filled with a material of dielectric constant K = 4. The equivalent capacitance of the system would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 32 μF
The capacitance of a capacitor filled with a dielectric is,
$C'=KC$ $[C_1=C=40μF]$
$⇒C'=4×40=160μF=C_2$
Now,
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$ [In series]
$\frac{1}{C_{eq}}=\frac{1}{40}+\frac{1}{160}$
$⇒C_{eq}=\frac{160}{5}=32μF$