**Answer:**

**Explanation:**

capacitance of capacitor

= ε₀ A / d , A is area of plate , d is plate separation

= (8.85 x10⁻¹² x 1.7² x 10⁻⁴) /( .35 x 10⁻³ )

= 73 x 10⁻¹³ F

Charge = 4.5 x 10⁻⁷

voltage between the plates,

Δ V = charge / capacitance

= 4.5 x 10⁻⁷ / 73 x 10⁻¹³

= .0616 x 10⁶

= 616 x 10² V

To make the vottage double , we will reduce capacitance to half .

To reduce capacitance half we will increase plate distance to double.

new plate distance

= 1.7 x 2

= 3.4 cm

**Answer:**

Hope it helps for you :)))))