91影视

The conductivity of the material is $蟽$.

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

The time constant is $蟿$.

The capacitance is $C$.

(a)

According to Gauss law,

$鈭�E路da=\frac{Q}{{鈭�}_0}$

The expression for the current density is given by,

$J=蟽E$

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

$l=鈭�Jda$

Substitute$蟽E$for J into above equation.

$$\begin{gathered} l=鈭�\left(蟽E\right)da \\ l=蟽鈭�Eda \end{gathered}$$

Substitute $\frac{Q}{{鈭�}_0}$for $鈭�E路da$into above equation.

$$\begin{gathered} l=蟽\frac{Q}{{鈭�}_0} \\ l=\frac{蟽Q}{{鈭�}_0} \end{gathered}$$

According to the ohm鈥檚 law

V = IR

Substitute $\frac{蟽Q}{{鈭�}_0}$for l into above equation.

$V=\frac{蟽Q}{{鈭�}_0}R$ 鈥︹�� (1)

The charge on the capacitor is given by,

Q = CV

Substitute$\frac{蟽Q}{{鈭�}_0}R$for V into above equation.

$$\begin{gathered} Q=C\frac{蟽Q}{{鈭�}_0}R \\ 1=C\frac{蟽Q}{{鈭�}_0}R \\ R=\frac{{鈭�}_0}{蟽Q} \end{gathered}$$

Hence the expression for the resistance$R=\frac{{鈭�}_0}{蟽Q}$ is obtained.

(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) .

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

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

$蟿=RC$

Substitute$\frac{{鈭�}_0}{蟽C}$ for R into above equation.

$$\begin{gathered} 蟿=\frac{{鈭�}_0}{蟽C}C \\ 蟿=\frac{{鈭�}_0}{蟽} \end{gathered}$$

Hence the expression for the time constant$蟿$ is $\frac{{鈭�}_0}{蟽}$.