Question

Two parallel plate capacitors, each of 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:

• A. 30 µF
• B. 31 µF
• C. 32 µF
• D. 33 µF

Answer 1

Answer: (C)

32 µF

Answer 2

$\textbf{Solution:} \text{ First, calculate the capacitance of the capacitor with the dielectric inserted. The new capacitance } C_1 \text{ with dielectric constant } K = 4 \text{ is:}$
$C_1 = K \times C = 4 \times 40 \, \mu\text{F} = 160 \, \mu\text{F}$
$\text{The second capacitor } C_2 \text{ remains unchanged at } 40 \, \mu\text{F}.$
$\text{The equivalent capacitance for capacitors in series is given by:}$
$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$
$\text{Substituting values:}$
$\frac{1}{C_{\text{eq}}} = \frac{1}{160 \, \mu\text{F}} + \frac{1}{40 \, \mu\text{F}} = \frac{1}{160} + \frac{1}{40} = \frac{1 + 4}{160} = \frac{5}{160}$
$C_{\text{eq}} = \frac{160}{5} = 32 \, \mu\text{F}$