Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.
The problem can be modeled through a linear equation, in the form:
With the initial conditions as,
Where Q(t) is the charge.
<em>The general solution of a linear equation is given as:</em>
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Applying this definiton in our differential equation we have that
To find b and a we use the first equation and find the roots:
Then we have
To find the values of the Constant we apply the initial conditions, then
And for the derivate:
We have a system of 2x2:
Solving we have:
The we can replace at the equation and we have that the Charge at any moment is given by,
If we obtain the derivate we find also the Current, then