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The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 12.0V (assuming circuit resistance is negligable) and it has a capacitance of 18.0μf or 18.0x10^-6f, therefore charge equals (18.0x10^-6)x12=2.16x10^-4C (Coulombs).
Answer:
Explanation:
Let initial extension in the spring= x₀
Force on the spring = F₀
Let spring constant = k
Fo = k x₀
Fn = 3k x₀
Fn /Fo = 3
PEs0 ( ORIGINAL) =1/2 k x₀²
PEsn ( NEW) =1/2 k (3x₀)²
PEsn / PEs0 = 9
