Answer
See answer below
Work Step by Step
Using RC Circuit we get:
$$\frac{dq}{dt}+\frac{q}{RC}=\frac{E}{R}$$
Since $E=0$
$$\frac{dq}{dt}+\frac{q}{RC}=0$$
The integrating factor is:
$$I=e^{\int \frac{1}{RC}dt}=e^{\frac{t}{RC}}$$
Multiplying by integrating factor:
$$e^{\frac{t}{RC}}\frac{dq}{dt}+e^{\frac{t}{RC}}\frac{q}{RC}=0$$
$$qe^{\frac{t}{RC}}=C$$
$$q=Ce^{\frac{-t}{RC}}$$
Since $q=5$ we get:
$$q=C=5 $$
Hence here,
$$q=5e^{\frac{-t}{RC}}$$
When $t \rightarrow \infty$:
$$\lim q=\lim 5e^{\frac{-t}{RC}}=0$$
and this is reasonable.
