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 dielectric material of dielectric constant K = 4. The equivalent capacitance of the system would be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 32 μF
Filling the space between the plates with a dielectric multiplies the capacitance by K. The two capacitors are then combined in series.
\( \displaystyle C_1 = 40\,\mu\mathrm{F} \qquad C_2 = KC = 4 \times 40\,\mu\mathrm{F} = 160\,\mu\mathrm{F} \)
\( \displaystyle C_{\mathrm{eq}} = \frac{C_1C_2}{C_1 + C_2} \)
\( \displaystyle C_{\mathrm{eq}} = \frac{40 \times 160}{40 + 160}\,\mu\mathrm{F} = \frac{6400}{200}\,\mu\mathrm{F} \)
\( \displaystyle C_{\mathrm{eq}} = 32\,\mu\mathrm{F} \)
Note: A similar question was also asked in CUET (UG) 2024 on 19-07-2024 (Slot 1).