Chapter 27, Problem 83PQ

(a)

To determine

The expression for energy stored in the capacitor $C_2$.

Answer to Problem 83PQ

The expression for energy stored in the capacitor $C_2$ is $U_2=\frac{U_1{C}_1{C}_2}{{\left(C_1+C_2\right)}^2}$.

Explanation of Solution

Write the expression to find the equivalent capacitance of the capacitors connected in parallel.

  $C_{eq}=C_1+C_2$                                                                                                        (I)

Here, $C_{eq}$ is the equivalent capacitance, $C_1$ is the capacitance of the first capacitor, $C_2$ is the capacitance of the second capacitor.

Write the expression charge on $C_1$.

  $Q=C_{1}V$                                                                                                              (II)

Here, $Q$ is the charge on $C_1$, $V$ is the voltage across $C_1$.

Write the expression to find the voltage across the parallel combination.

  $V_n=\frac{Q}{C_{eq}}$                                                                                                             (III)

Here, $V_n$ is the voltage across the parallel combination.

Substitute equation (I), (II) in (III) to find the voltage across the parallel combination.

  $V_n=\frac{C_{1}V}{C_1+C_2}$

Write the equation for initial energy in the capacitor $C_1$.

  $U_1=\frac{1}{2}{C}_1{V}^2$                                                                                                             (IV)

Here, $U_1$ is the initial energy in the capacitor $C_1$.

Write the equation for final energy in the capacitor $C_2$.

  $U_2=\frac{1}{2}{C}_2{V}_n^2$

Here, $U_2$ is the initial energy in the capacitor $C_2$.

Substitute $\frac{C_{1}V}{C_1+C_2}$ for $V_n$ to find the final energy in the capacitor $C_2$.

  $\begin{array}{c}{U}_2=\frac{1}{2}{C}_2{\left(\frac{C_{1}V}{C_1+C_2}\right)}^2\\ =\frac{1}{2}{C}_2\frac{C_1^2{V}^2}{C_1^2+C_2^2}\\ =\left(\frac{1}{2}{C}_1{V}^2\right)\left(\frac{C_1{C}_2}{{\left(C_1+C_2\right)}^2}\right)\end{array}$

Conclusion:

Re-write the expression using equation (IV).

  $\begin{array}{c}{U}_2=U_1\left(\frac{C_1{C}_2}{{\left(C_1+C_2\right)}^2}\right)\\ =\frac{U_1{C}_1{C}_2}{{\left(C_1+C_2\right)}^2}\end{array}$

Therefore, the expression for energy stored in the capacitor $C_2$ is $U_2=\frac{U_1{C}_1{C}_2}{{\left(C_1+C_2\right)}^2}$.

(b)

To determine

Show that $U_2$ is maximum when $C_1=C_2$.

Answer to Problem 83PQ

$U_2$ is maximum when $C_1=C_2$.

Explanation of Solution

Write the expression for $U_2$ from part (a).

  $U_2=\frac{U_1{C}_1{C}_2}{{\left(C_1+C_2\right)}^2}$

Re-arrange the above expression to find the maximum point for $U_2$.

  $\begin{array}{c}{U}_2=\frac{U_1{C}_1{C}_2}{{\left(C_1+C_2\right)}^2}\\ =\frac{1}{\frac{C_1}{C_2}+2+\frac{C_2}{C_1}}{U}_1\end{array}$

Substitute $x$ for $\frac{C_1}{C_2}$ to find the maximum point for $U_2$.

  $U_2=\frac{1}{x+2+\frac{1}{x}}{U}_1$

The denominator has to be minimum to maximize the $U_2$. The value of $x+2+\frac{1}{x}$ will be minimum when the derivative of $x+2+\frac{1}{x}$ is equal to zero.

Differentiate $x+2+\frac{1}{x}$ with respect to $x$ and equate to zero to find the maximum point for $U_2$.

  $\begin{array}{c}\frac{d}{dx}\left(x+2+\frac{1}{x}\right)=0\\ 1-\frac{1}{x^2}=0\\ x^2=1\\ x=\pm 1\end{array}$

Differentiate the above expression to find the points of maxima and minima

  $\frac{d}{dx}\left(1-\frac{1}{x^2}\right)=\frac{2}{x^3}$

The second derivative gives a positive value when $x=1$. Thus, the minimum point of the function is at $x=1$. Since, $x=\frac{C_1}{C_2}=1$; $C_1=C_2$.