Solution:

The capacitance of a parallel-plate capacitor is given by the general formula:

$$C = \frac{\varepsilon_0 A}{d}$$

where:

• 
  $C$ 

  = capacitance,

• 
  $\varepsilon_0$ 

  = permittivity of free space,

• 
  $A$ 

  = effective overlapping area of the two plates,

• 
  $d$ 

  = separation between the plates.

For Plates of Unequal Areas ( $A_1 < A_2$

):

When two plates of areas

$A_1$

and

$A_2$

are used to form a capacitor, only the smaller area (

$A_1$

) determines the effective overlapping area for charge storage. This is because the excess area of the larger plate (

$A_2 - A_1$

) does not contribute to the capacitance.

Thus, the capacitance is:

$$C = \frac{\varepsilon_0 A_1}{d}$$

Final Answer:

The capacitance of the capacitor is:

$$\boxed{C = \frac{\varepsilon_0 A_1}{d}}$$