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.
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