Question

(a) Two metal objects are embedded in weakly conducting material of conductivity $\mathit{\sigma }$(Fig. 7 .6). Show that the resistance between them is related to the capacitance of the arrangement by

$\mathbf{R}\mathbf{=}\frac{{\mathbf{\in }}_{\mathbf{0}}}{\mathbf{\sigma }\mathbf{C}}$

(b) Suppose you connected a battery between 1 and 2, and charged them up to a potential difference${\mathit{V}}_{\mathbf{0}}$. If you then disconnect the battery, the charge will gradually leak off. Show that$\mathbf{V}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{r}}$, and find the time constant,$\mathit{\tau }$, in terms of ${\mathbf{\in }}_{\mathbf{0}}$and .$\mathit{\sigma }$

Short answer

(a) The expression for the resistance is obtained $R=\frac{{\in }_0}{\sigma C}$.

(b) The expression for the voltage$V\left(t\right)=V_0{e}^{-\frac{t}{RC}}$ is obtained and the expression for the time constant$\tau$ is $\frac{{\in }_0}{C}$.

Step-by-step solution

Write the given data from the question.

The conductivity of the material is $\sigma$.

The potential difference between the 1 and 2 is $V_0$.

The time constant is $\tau$.

The capacitance is $C$.

Determine the expression for the resistance.

(a)

According to Gauss law,

$\int E·da=\frac{Q}{{\in }_0}$

The expression for the current density is given by,

$J=\sigma E$

The relationship between the current l and current density J is given by,

$l=\int Jda$

Substitute$\sigma E$for J into above equation.

$$\begin{gathered} l=\int \left(\sigma E\right)da \\ l=\sigma \int Eda \end{gathered}$$

Substitute $\frac{Q}{{\in }_0}$for $\int E·da$into above equation.

$$\begin{gathered} l=\sigma \frac{Q}{{\in }_0} \\ l=\frac{\sigma Q}{{\in }_0} \end{gathered}$$

According to the ohm’s law

V = IR

Substitute $\frac{\sigma Q}{{\in }_0}$for l into above equation.

$V=\frac{\sigma Q}{{\in }_0}R$ …… (1)

The charge on the capacitor is given by,

Q = CV

Substitute$\frac{\sigma Q}{{\in }_0}R$for V into above equation.

$$\begin{gathered} Q=C\frac{\sigma Q}{{\in }_0}R \\ 1=C\frac{\sigma Q}{{\in }_0}R \\ R=\frac{{\in }_0}{\sigma Q} \end{gathered}$$

Hence the expression for the resistance$R=\frac{{\in }_0}{\sigma Q}$ is obtained.

Determine the expression for the voltage and time constant.

(b)

The charge on the capacitor is given by,

Q = CV

Substitute IR for V into above equation.

$$\begin{gathered} Q=CIR \\ I=\frac{Q}{RC} \end{gathered}$$

Here is l positive, therefore the charge of capacitor is decreasing.

The current is defined as the rate of charge of moving charge.

$\frac{dQ}{dt}=-1$

Substitute$\frac{Q}{RC}$ for l into above equation,

$\frac{dQ}{dt}=-\frac{Q}{RC}$

By integrating the above equation from$Q_0$ to Q(t) .

$$\begin{gathered} {\int }_Q^{Q^{\left(t\right)}}\frac{dq}{Q}={\int }_0^t-\frac{dt}{RC} \\ \mathrm{In}{\left(Q\right)}_{Q_0}^{Q\left(t\right)}=-\frac{1}{RC}{\left(t\right)}_0^t+\mathrm{InA} \\ \mathrm{In}\left(\mathrm{Q}\right(\mathrm{t})-{\mathrm{Q}}_0)=-\frac{1}{\mathrm{RC}}(\mathrm{t}-0)+\mathrm{InA} \\ \mathrm{In}\left(\frac{\mathrm{Q}\left(\mathrm{t}\right)}{{\mathrm{Q}}_0}\right)=-\frac{1}{\mathrm{RC}}\left(\mathrm{t}\right)+\mathrm{InA} \end{gathered}$$ …… (2)

Here is A the integration constant.

At t = 0,Q(t) = 0

$$\begin{gathered} \mathrm{In}\left(\frac{\mathrm{Q}\left(0\right)}{{\mathrm{Q}}_0}\right)=-\frac{1}{\mathrm{RC}}\left(0\right)+\mathrm{InA} \\ \mathrm{In}\left(\frac{\left(0\right)}{{\mathrm{Q}}_0}\right)=-\frac{1}{\mathrm{RC}}\left(0\right)+\mathrm{InA} \end{gathered}$$

In A = 0

Substitute 0 or In A into equation (2)

$$\begin{gathered} \mathrm{In}\left(\frac{\mathrm{Q}\left(\mathrm{t}\right)}{{\mathrm{Q}}_0}\right)=-\frac{1}{\mathrm{RC}}\left(t\right)+0 \\ \left(\frac{\mathrm{Q}\left(\mathrm{t}\right)}{{\mathrm{Q}}_0}\right)=e^{-\frac{1}{\mathrm{RC}}\left(t\right)} \\ Q\left(t\right)=Q_0{e}^{{}^{-\frac{1}{\mathrm{RC}}\left(t\right)}} \end{gathered}$$ …… (3)

The expression for the initial charge is given by,

$Q_0=C{V}_0$

The expression for the voltage at time t is given by,

$Q\left(t\right)=CV\left(t\right)$

Substitute CV(t) for Q(t) and$C{V}_0$ for$Q_0$ into equation (3).

$$\begin{gathered} CV\left(t\right)=C{V}_0{e}^{-\frac{t}{RC}} \\ V\left(t\right)=V_0{e}^{-\frac{t}{RC}} \end{gathered}$$

Hence the expression for the voltage is obtained.

$V\left(t\right)=V_0{e}^{-\frac{t}{RC}}$

The time constant is given by,

$\tau =RC$

Substitute$\frac{{\in }_0}{\sigma C}$ for R into above equation.

$$\begin{gathered} \tau =\frac{{\in }_0}{\sigma C}C \\ \tau =\frac{{\in }_0}{\sigma } \end{gathered}$$

Hence the expression for the time constant$\tau$ is $\frac{{\in }_0}{\sigma }$.