On connecting a 9V battery to a **Capacitor** consisting of two circular plates of radius 0.066 m separated by an air gap of 2. 0 mm. The charge on the positive plate is **544.7 ×10⁻¹² C**.

**Capacitance** of the capacitor is determined by the area of the plate of Capacitor and distance between the plates of capacitor.

Let the area of the **Capacitor** be A , radius of circular plates be r and distance between the plates of capacitor be d.

Given, Voltage, V = 9V

Radius, r = 0.066m

Distance, d = 2mm = 0.002m

Area of the Capacitor, A = πr²

A = π(0.066)²

A = 0.013m²

Capacitance, C = ε₀A / d

C = 8.85×10⁻¹²×0.013/0.002

C = 60.5 ×10⁻¹² F

C = 60.5 pF

We know that Q = CV where Q is the charge on capacitor.

Q = 60.5 ×10⁻¹² × 9

Q= 544.7 ×10⁻¹² C

Since, both plates of a capacitor acquire equal and opposite charge.

Hence the charge on the positive plate of the capacitor is **544.7 ×10⁻¹² C**.

Learn more about **Capacitance** here, brainly.com/question/14746225

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