Question

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?

Short answer

The value of time on each day is 96.26h and on the humid days0.048517 sec .

Step-by-step solution

Important formula.

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

Evaluate the value of q

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

Find the value of time on each day

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

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

Calculate the value of time in humid days.

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 .