Chapter 25, Problem 99P

(a)

To determine

The charge on the capacitor plate.

Answer to Problem 99P

The charge on the capacitor plate is $5.69\mu \text{C}$ .

Explanation of Solution

Given:

The value of emf is $\epsilon =6.00\text{V}$ .

The value of capacitance is $C=1.50\mu \text{F}$ .

The resistance is $R=2.00\text{M}\Omega$ .

Formula used:

The expression for the charge as a function of time is given as,

  $Q\left(t\right)=C\epsilon \left(1-e^{-t/\tau }\right)$

Calculation:

The charge on the capacitor plate after $t=\tau \sec$ is calculated as,

  $\begin{array}{c}Q\left(t\right)=\left(1.50\mu \text{F}\right)\left(6.00\text{V}\right)\left(1-e^{-\tau /\tau }\right)\\ =\left(1.50\mu \text{F}\right)\left(6.00\text{V}\right)\left(1-e^{-1}\right)\\ =5.689\mu C\\ \approx 5.69\mu C\end{array}$

Conclusion:

Therefore, the charge on the capacitor plate is $5.69\mu \text{C}$ .

(b)

To determine

The rate of increasing charge.

Answer to Problem 99P

The rate of increasing charge is $1.10\mu \text{C}/\text{s}$ .

Explanation of Solution

Formula used:

The expression for rate of increasing charge is,

  $\frac{dQ}{dt}=\frac{\epsilon }{R}{e}^{-t/\tau }$

Calculation:

The rate of increasing charge after $t=\tau \sec$ is calculated as,

  $\begin{array}{c}\frac{dQ}{dt}=\left(\frac{6.00\text{V}}{2.00\text{M}\Omega }\right)e^{-\tau /\tau }\\ =\left(\frac{6.00\text{V}}{2.00\times {10}^6\Omega }\right)e^{-1}\\ =1.104\times {10}^{-6}\text{C}/\text{s}\\ =1.10\mu \text{C}/\text{s}\end{array}$ ..... (1)

Conclusion:

Therefore, the rate of increasing charge is $1.10\mu \text{C}/\text{s}$ .

(c)

To determine

The current after $t=\tau \sec$ .

Answer to Problem 99P

The current after $t=\tau \sec$ is $1.10\mu \text{A}$ .

Explanation of Solution

Formula used:

The expression for current is,

  $I\left(t\right)=\frac{dQ}{dt}$ ..... (2)

Calculation:

From equation (1) and (2), the current after $t=\tau \sec$ is calculated as,

  $\begin{array}{c}I\left(t\right)=1.10\mu \text{C}/\text{s}\\ =1.10\mu \text{A}\end{array}$

Conclusion:

Therefore, the current after $t=\tau \sec$ is $1.10\mu \text{A}$ .

(d)

To determine

The power supplied by battery.

Answer to Problem 99P

The power supplied by battery is $6.62\mu \text{W}$ .

Explanation of Solution

Formula used:

The expression for power is,

  $P\left(t\right)=\epsilon \cdot I\left(t\right)$

Calculation:

The power supplied by battery after $t=\tau \sec$ is calculated as,

  $\begin{array}{c}P\left(t\right)=\left(6.00\text{V}\right)\left(1.104\mu \text{A}\right)\\ =6.62\mu \text{W}\end{array}$

Conclusion:

Therefore, the power supplied by battery is $6.62\mu \text{W}$ .

(e)

To determine

The power delivered to resistor.

Answer to Problem 99P

The power delivered to resistor is $2.44\mu \text{W}$ .

Explanation of Solution

Formula used:

The expression for power delivered to resistor is,

  $P_R\left(t\right)=I^2\left(t\right)\cdot R$

Calculation:

The power supplied by battery after $t=\tau \sec$ is calculated as,

  $\begin{array}{c}{P}_R\left(t\right)={\left(1.104\mu \text{A}\right)}^2\left(2.00\text{M}\Omega \right)\\ ={\left(1.104\times {10}^{-6}\text{A}\right)}^2\left(2.00\times {10}^6\Omega \right)\\ =2.437\mu \text{W}\\ \text{=2}\text{.44}\mu \text{W}\end{array}$

Conclusion:

Therefore, the power delivered to resistor is $2.44\mu \text{W}$ .

(f)

To determine

The rate of stored energy.

Answer to Problem 99P

The rate of stored energy is $4.19\mu \text{W}$ .

Explanation of Solution

Formula used:

The expression for stored energy is,

  $U\left(t\right)=\frac{1}{2C}{Q}^2\left(t\right)$ ..... (3)

Calculation:

Differentiate equation (3) with respect to time,

  $\begin{array}{c}\frac{dU\left(t\right)}{dt}=\frac{1}{2C}\left\{2Q\left(t\right)\right\}\frac{dQ\left(t\right)}{dt}\\ =\frac{Q\left(t\right)}{C}I\left(t\right)\end{array}$

The rate of stored energy is calculated as,

  $\begin{array}{c}\frac{dU\left(t\right)}{dt}=\frac{5.69\mu \text{C}}{1.50\mu \text{F}}\left(1.104\mu \text{A}\right)\\ =4.187\mu \text{W}\\ =4.19\mu \text{W}\end{array}$

Conclusion:

Therefore, the rate of stored energy is $4.19\mu \text{W}$ .