Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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, https://brainly.com/question/14746225
#SPJ4
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
