When a charge of 1.850×10-4 C is applied to plates of a capacitor?
When a charge of 1.850×10-4 C is applied to the plates of a capacitor, the potential difference between the plates is 17 Volts. What is the capacitance?
I need the answer and a formula please...not just the law....
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Simple; Capacitance = Charge/Potential difference...
C = Q/V...
C = 1.850 x 10^-4 / 17
So C = 1.088 x 10^-5 Farads