Chapter 18, Problem 134P

(a)

To determine

Find the charge on the upper plate after the capacitor fully charged.

Answer to Problem 134P

The charge on the upper plate after the capacitor fully charged is $\text{9}\text{.9 nC}$.

Explanation of Solution

Write the equation for charge on upper plate.

$Q=CV=\frac{{\epsilon }_{0}A}{d}V=\frac{{\epsilon }_0{L}^2}{d}V$ (I)

Here, $Q$ is the charge on upper plate, $V$ is potential difference, ${\epsilon }_0$ permittivity of free space, $L$ is length of the plate and $d$ is the distance between the plates.

Conclusion:

Substitute $10.0\text{ V}$ for $V$, $8.854\times {10}^{-12}{\text{ C}}^2\text{/N}{\text{.m}}^2$ for ${\epsilon }_0$, $0.10\text{ m}$ for $L$ and $89\times {10}^{-6}\text{ m}$ for $d$ in equation I.

$\begin{array}{c}Q=\frac{\left(8.854\times {10}^{-12}{\text{ C}}^2\text{/N}{\text{.m}}^2\right){\left(0.10\text{ m}\right)}^2\left(10.0\text{ V}\right)}{89\times {10}^{-6}\text{ m}}\\ =9.9\text{ nC}\end{array}$

Therefore, the charge on the upper plate after the capacitor fully charged is $\text{9}\text{.9 nC}$.

(b)

To determine

Draw the graph between current through the resistor with time.

Answer to Problem 134P

The graph between current through the resistor with time is given below.

Explanation of Solution

Write the equation for time constant.

$\tau =RC$ (II)

Here, $\tau$ is the time constant and $R$ is the resistance.

Substitute the value of $C$ from equation I in equation II.

$\tau =R\frac{{\epsilon }_0{L}^2}{d}$ (III)

Write the equation for initial current.

$I_0=\frac{V}{R}$ (IV)

Write the equation for variation of current with time.

$I=I_0{e}^{-t/\tau }$ (V)

Conclusion:

Substitute $0.100\times {10}^6\Omega$ for $R$, $8.854\times {10}^{-12}{\text{ C}}^2\text{/N}{\text{.m}}^2$ for ${\epsilon }_0$, $0.10\text{ m}$ for $L$ and $89\times {10}^{-6}\text{ m}$ for $d$ in equation III.

$\begin{array}{c}\tau =\frac{\left(0.100\times {10}^6\Omega \right)\left(8.854\times {10}^{-12}{\text{ C}}^2\text{/N}{\text{.m}}^2\right){\left(0.10\text{ m}\right)}^2}{89\times {10}^{-6}\text{ m}}\\ =9.9\times {10}^{-5}\text{ s}\end{array}$

Substitute $10.0\text{ V}$ for $V$ and $0.100\times {10}^6\Omega$ for $R$ in equation IV.

$I_0=\frac{10.0\text{ V}}{0.100\times {10}^6\Omega }=100\text{ μA}$

Substitute $100\text{ μA}$ for $I_0$ and $9.9\times {10}^{-5}\text{ s}$ for $\tau$ in equation V.

$I=\left(100\text{ μA}\right)e^{-t/\left(9.9\times {10}^{-5}\text{ s}\right)}$

Therefore, the graph between current through the resistor with time using the above equation is given below.

(b)

To determine

Find the energy dissipated in whole discharging process.

Answer to Problem 134P

The energy dissipated in whole discharging process is $50\text{ nJ}$.

Explanation of Solution

Write the equation for total energy.

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{{\epsilon }_0{L}^2}{d}\right)V^2$ (VI)

Here, $U$ is the total energy.

Conclusion:

Substitute $10.0\text{ V}$ for $V$, $8.854\times {10}^{-12}{\text{ C}}^2\text{/N}{\text{.m}}^2$ for ${\epsilon }_0$, $0.10\text{ m}$ for $L$ and $89\times {10}^{-6}\text{ m}$ for $d$ in equation III.

$\begin{array}{c}U=\frac{1}{2}\frac{\left(8.854\times {10}^{-12}{\text{ C}}^2\text{/N}{\text{.m}}^2\right){\left(0.10\text{ m}\right)}^2{\left(10.0\text{ V}\right)}^2}{89\times {10}^{-6}\text{ m}}\\ =50\text{ nJ}\end{array}$

Therefore, the energy dissipated in whole discharging process is $50\text{ nJ}$.