Now solve the differential equation V(t)=−CRdV(t)dt for the initial conditions given in the problem introduction to find the voltage as a function of time for any time t. Express your answer in terms of q0, C, R, and t.

V(t) =(q0/C) * e^(−t/RC )

Explanation:

If there were a battery in the circuit with EMF  E , the equation for  V(t)  would be  V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in  V(t)  (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution  Vb(t)=V(t)−E . The effect of this substitution is to eliminate the  E  term and yield an equation for  Vb(t)  that is identical to the equation you solved for  V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time  t=0 . The solution under these conditions will look like  V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time  t=0  (since the capacitor was uncharged then) and rises asymptotically to  E  (with the result that current essentially stops flowing through the circuit).

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