Answer:
"Becomes four times
"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.