Solution:

To solve this, we determine the combination of capacitors required to achieve the desired capacitance and voltage.

Given:

• Individual capacitor:
  $C = 8 \, \mu\text{F}$ 

  , $V_{\text{max}} = 250 \, \text{V}$ 

• Desired combination:
  $C_{\text{req}} = 16 \, \mu\text{F}$ 

  , $V_{\text{req}} = 1000 \, \text{V}$ 

Step 1: Voltage requirement

To achieve

$V_{\text{req}} = 1000 \, \text{V}$

, multiple capacitors must be connected in series because the voltage across a series combination adds up. The number of capacitors required in series is:

$$n = \frac{V_{\text{req}}}{V_{\text{max}}} = \frac{1000}{250} = 4$$

Thus, 4 capacitors in series are required to handle 1000 V.

Step 2: Capacitance in series

The effective capacitance of

$n$

capacitors in series is given by:

$$C_{\text{series}} = \frac{C}{n} = \frac{8}{4} = 2 \, \mu\text{F}$$

So, a series of 4 capacitors provides

$C_{\text{series}} = 2 \, \mu\text{F}$

.

Step 3: Capacitance requirement

To achieve

$C_{\text{req}} = 16 \, \mu\text{F}$

, multiple such series groups must be connected in parallel because capacitance in parallel adds up. The number of such series groups required is:

$$m = \frac{C_{\text{req}}}{C_{\text{series}}} = \frac{16}{2} = 8$$

Thus, 8 series groups are required.

Step 4: Total capacitors

Each series group contains 4 capacitors, and there are 8 such groups. Therefore, the total number of capacitors is:

$$\text{Total capacitors} = n \cdot m = 4 \cdot 8 = 32$$

Final Answer:

The minimum number of capacitors required is:

$$\boxed{32}$$