A 10^-8-Fcapacitor (10 nano-farads) is charged to 50Vand then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is$\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{\Omega }$; on a humid day, the resistance is$\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\Omega }$. How long will it take the capacitor voltage to dissipate to half its original value on each day?

The equation is$\mathbf{R}\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{0}$.

Here, $\mathbf{C}\mathbf{=}{\mathbf{10}}^{\mathbf{8}}\mathbf{\Omega }$, q = 50V,$\mathbf{r}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\Omega }$

The equation is

$$\begin{gathered} \mathbf{R}\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{0} \\ \mathbf{R}\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{q}}{\mathbf{C}} \\ \frac{\mathbf{dq}}{\mathbf{q}}\mathbf{=}\mathbf{-}\frac{\mathbf{dt}}{\mathbf{CR}} \\ \mathbf{lnq}\mathbf{=}\mathbf{-}\frac{\mathbf{t}}{\mathbf{CR}}\mathbf{+}\mathbf{K} \\ \mathbf{q}\mathbf{=}{\mathbf{Ke}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{CR}}} \end{gathered}$$

When $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{q}\mathbf{=}\mathbf{50} \mathbf{V}\mathbf{,} \mathbf{then} \mathbf{K}\mathbf{=}\mathbf{50}$

role="math" localid="1664228869574" $\mathbf{q}\mathbf{=}\mathbf{50}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{C}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{r}\mathbf{)}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left(\mathrm{Where},r\mathrm{is}\mathrm{the}\mathrm{resistance}\mathrm{of}\mathrm{the}\mathrm{air}\mathrm{gap}\right)$

The equation for the capacitor voltage

$$\begin{gathered} {\mathbf{V}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{q}}{\mathbf{C}} \\ {\mathbf{V}}_{\mathbf{C}}\mathbf{=}\mathbf{-}\frac{\mathbf{50}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{C}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{r}\mathbf{)}}}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}} \\ {\mathbf{V}}_{\mathbf{C}}\mathbf{=}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{5}\mathbf{\times }\mathbf{1}{\mathbf{0}}^{\mathbf{13}}\mathbf{)}}} \end{gathered}$$

Since the capacitor voltage to dissipate to half its original value each day.

So, the equation will be

$$\begin{gathered} {\mathbf{V}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{V}}_{{\mathbf{C}}_{\mathbf{o}}} \\ \mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{5}\mathbf{\times }\mathbf{1}{\mathbf{0}}^{\mathbf{13}}\mathbf{)}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{50}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{0}}{\mathbf{C}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{r}\mathbf{)}}}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}} \\ \mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{5}\mathbf{\times }\mathbf{1}{\mathbf{0}}^{\mathbf{13}}\mathbf{)}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

When R = 0 then

$$\begin{gathered} {\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{(}\mathbf{0}\mathbf{+}\mathbf{5}\mathbf{\times }\mathbf{1}{\mathbf{0}}^{\mathbf{13}}\mathbf{)}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \\ {\mathbf{lne}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}}}\mathbf{=}\mathbf{ln}\frac{\mathbf{1}}{\mathbf{2}} \\ \mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6931}\mathbf{\times }\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}} \\ \mathbf{t}\mathbf{=}\mathbf{346560}\mathbf{sec} \\ \mathbf{t}\mathbf{=}\mathbf{96}\mathbf{.}\mathbf{26}\mathbf{h} \end{gathered}$$

Here, $\mathbf{r}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\Omega }$, and R = 0 then

$$\begin{gathered} {\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{(}\mathbf{0}\mathbf{+}\mathbf{7}\mathbf{\times }\mathbf{1}{\mathbf{0}}^{\mathbf{6}}\mathbf{)}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \\ {\mathbf{lne}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}}\mathbf{=}\mathbf{ln}\frac{\mathbf{1}}{\mathbf{2}} \\ \mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6931}\mathbf{\times }\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}} \\ \mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{048517}\mathbf{sec} \end{gathered}$$

Therefore, the value of time on each day is 96.26h and on the humid days 0.048517 sec .