Time required to charge a capacitor

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To determine the time required to charge a capacitor with a capacitance of 28.7 μF to a voltage of 28.9 V using a constant current of 63.2 nA, the charge on the capacitor can be calculated using the formula Q = CV. The charge at 28.9 V is approximately 0.000829 C. Since the current is constant, the time can be found using the relationship t = Q/I, resulting in a time of about 13,141 seconds or approximately 3.65 hours. The exponential charging equation is unnecessary in this scenario due to the constant current. Understanding these calculations is crucial for solving similar capacitor charging problems.
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Homework Statement


A capacitor with a capacitance of C = 28.7 μF is slowly charged by a constant current of I = 63.2 nA. How long does it take to charge the capacitor to a voltage of V = 28.9 V?


Homework Equations


q=cvb(1-e^(-t/RC))
Vb=IR+QC


The Attempt at a Solution


I have tried to solve for R by R=I/V but I don't know how to solve for t because Q is not given, and I don't know how to find it. Any ideas? I only got a couple hours left. Any help is greatly appreciated.
 
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The statement that the current is constant says that you are in the linear region all the way to charging the capacitor to 28.9 V. You can calculate how much charge is on the plates when the voltage is 28.9 V and from this the time it takes for this charge to accumulate at a constant rate. The exponential is not needed here.
 

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