SOLUTION

Step 1: Capacitance Formula

$C = \frac{\kappa \epsilon_0 A}{d}$

For air as the dielectric:

$C = \frac{\epsilon_0 A}{d}$

Step 2: Initial Conditions

– Initial area: $ A_1 = A $
– Initial distance: $ d_1 = d $
– Initial capacitance:

$C_1 = \frac{\epsilon_0 A}{d}$

Step 3: New Conditions

– New area: $ A_2 = \frac{A}{2} $
– New distance: $ d_2 = 2d $
– New capacitance:
$C_2 = \frac{\epsilon_0 A_2}{d_2} = \frac{\epsilon_0 \left(\frac{A}{2}\right)}{2d}$

Step 4: Simplifying New Capacitance

$C_2 = \frac{\epsilon_0 \cdot \frac{A}{2}}{2d} = \frac{\epsilon_0 A}{4d}$

Step 5: Relationship Between Voltage and Capacitance

Voltage across a capacitor:

$V = \frac{Q}{C}$

Initial voltage:

$
V_1 = \frac{Q}{C_1} = \frac{Q}{\frac{\epsilon_0 A}{d}} = \frac{Q d}{\epsilon_0 A}$

New voltage:
$V_2 = \frac{Q}{C_2} = \frac{Q}{\frac{\epsilon_0 A}{4d}} = \frac{4Q d}{\epsilon_0 A}$

Step 6: Comparing Voltages

$V_2 = 4V_1$

Conclusion
When the distance between the plates is doubled and the effective area of the plates is made half, the voltmeter reading becomes four times greater than before.