Update Capacitance of $ C_1 $:

The initial capacitance of $ C_1 $ is $ 5 \, \mu \text{F} $. Adding a dielectric with a dielectric constant of 4 increases $ C_1 $ to:

$ C_1 = 4 \times 5 = 20 \, \mu \text{F} $

Identify Capacitances of $ C_2 $ and $ C_3 $:

Both $ C_2 $ and $ C_3 $ remain at $ 5 \, \mu \text{F} $ since no dielectric is added to them.

$ C_2 = 5 \, \mu \text{F}, \quad C_3 = 5 \, \mu \text{F} $

Calculate Effective Capacitance:

Capacitors $ C_1 $ and $ C_2 $ are in series, and this series combination is parallel to $ C_3 $.

Series Combination of $ C_1 $ and $ C_2 $:

The formula for capacitors in series is:

$ C_{\text{series}} = \frac{C_1 \times C_2}{C_1 + C_2} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4 \, \mu \text{F} $

Parallel Combination with $ C_3 $:

The formula for capacitors in parallel is:

$ C_{\text{eq}} = C_{\text{series}} + C_3 = 4 + 5 = 9 \, \mu \text{F} $

Thus, the effective capacitance between points $ A $ and $ B $ is $ 9 \, \mu \text{F} $.