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Sagot :
Charge appeared on the plates of a 6.0 μf capacitor when it is charged to 40V is 240μC.
Capacitance of the capacitor is determined by the area of the plate of Capacitor and distance between the plates of capacitor.
Capacitor is dependent on the area of the plates, distance between the plates and medium present between the plates.
It is not dependent of the charge and voltage across the capacitor.
We know that Q = CV, where Q is the charge on capacitor, C is the Capacitance and V is the voltage applied across the capacitor.
Q = 6 × 10⁻⁶ × 40
Q = 240× 10⁻⁶ C
Q = 240 μC
Hence the charge on the plates of the capacitor is 240 μC.
Learn more about Capacitance here, brainly.com/question/14746225
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