Chapter 26, Problem 40P

(a)

To determine

To show: The energy associated with a single conducting sphere is $U_{\text{E}}=\frac{k_{\text{e}}{q}^2}{2R}$.

Answer to Problem 40P

The energy associated with a single conducting sphere is $U_{\text{E}}=\frac{k_{\text{e}}{q}^2}{2R}$.

Explanation of Solution

Given info: The radii of two conducting sphere is $R_1$ and $R_2$. The total charge shared between them is $Q$. The charge on first and second sphere is $q_1$ and $q_2$ respectively.

Write the expression to calculate the capacitance of a sphere of radius $R$.

$C=\frac{R}{k_{\text{e}}}$

Here,

$k_{\text{e}}$ is the Coulomb’s law constant.

Write the expression to calculate the potential difference.

$\Delta V=\frac{k_{\text{e}}q}{R}$

Here,

$\Delta V$ is the potential difference of the capacitor.

$q$ is the charge on a single sphere.

Write the expression to calculate the energy stored in the capacitor.

$U_{\text{E}}=\frac{1}{2}C{\left(\Delta V\right)}^2$

Substitute $\frac{R}{k_{\text{e}}}$ for $C$ and $\frac{k_{\text{e}}q}{R}$ for $\Delta V$ in above equation.

$\begin{array}{c}{U}_{\text{E}}=\frac{1}{2}\left(\frac{R}{k_{\text{e}}}\right){\left(\frac{k_{\text{e}}q}{R}\right)}^2\\ =\frac{k_{\text{e}}{q}^2}{2R}\end{array}$

Conclusion:

Therefore, the energy associated with a single conducting sphere is $U_{\text{E}}=\frac{k_{\text{e}}{q}^2}{2R}$.

(b)

To determine

The total energy of the system of two spheres in term of $q_1$, $Q$ and the radii $R_1$ and $R_2$.

Answer to Problem 40P

The total energy of the system of two spheres in term of $q_1$, $Q$ and the radii $R_1$ and $R_2$ is $\frac{1}{2}\frac{k_{\text{e}}{q}_1^2}{R_1}+\frac{k_{\text{e}}}{2}\frac{{\left(Q-q_1\right)}^2}{R_2}$.

Explanation of Solution

Given info: The radii of two conducting sphere is $R_1$ and $R_2$. The total charge shared between them is $Q$. The charge on first and second sphere is $q_1$ and $q_2$ respectively.

Write the expression to calculate the capacitance of a sphere of radius $R$.

$C=\frac{R}{k_{\text{e}}}$

Write the expression to calculate the total energy of the system of two sphere.

$U_{\text{E}}=\frac{1}{2}\frac{q_1^2}{C_1}+\frac{1}{2}\frac{q_2^2}{C_2}$

Substitute $\frac{R_1}{k_{\text{e}}}$ for $C_1$ and $\frac{R_2}{k_{\text{e}}}$ for $C_2$ in above equation.

$U_{\text{E}}=\frac{1}{2}\frac{q_1^2}{\left(\frac{R_1}{k_{\text{e}}}\right)}+\frac{1}{2}\frac{q_2^2}{\left(\frac{R_2}{k_{\text{e}}}\right)}$ (1)

The sum of charge of both sphere are,

$\begin{array}{c}Q=q_1+q_2\\ q_2=Q-q_1\end{array}$

Substitute $Q-q_1$ for $q_2$ in above equation.

$U_{\text{E}}=\frac{1}{2}\frac{k_{\text{e}}{q}_1^2}{R_1}+\frac{k_{\text{e}}}{2}\frac{{\left(Q-q_1\right)}^2}{R_2}$

Thus, the total energy of the system of two spheres in term of $q_1$, $Q$ and the radii $R_1$ and $R_2$ is $\frac{1}{2}\frac{k_{\text{e}}{q}_1^2}{R_1}+\frac{k_{\text{e}}}{2}\frac{{\left(Q-q_1\right)}^2}{R_2}$.

Conclusion:

Therefore, the total energy of the system of two spheres in term of $q_1$, $Q$ and the radii $R_1$ and $R_2$ is $\frac{1}{2}\frac{k_{\text{e}}{q}_1^2}{R_1}+\frac{k_{\text{e}}}{2}\frac{{\left(Q-q_1\right)}^2}{R_2}$.

(c)

To determine

The value of $q_1$ by differentiating the result of part (b).

Answer to Problem 40P

The value of $q_1$ is $q_1=\frac{R_{1}Q}{R_1+R_2}$.

Explanation of Solution

Given info: The radii of two conducting sphere is $R_1$ and $R_2$. The total charge shared between them is $Q$. The charge on first and second sphere is $q_1$ and $q_2$ respectively.

The total energy of the system of two spheres in term of $q_1$, $Q$ and the radii $R_1$ and $R_2$ is,

$U_{\text{E}}=\frac{1}{2}\frac{k_{\text{e}}{q}_1^2}{R_1}+\frac{k_{\text{e}}}{2}\frac{{\left(Q-q_1\right)}^2}{R_2}$.

Differentiate the above equation with respect to $q_1$ and equate to zero.

$\begin{array}{c}\frac{d{U}_{\text{E}}}{d{q}_1}=0\\ \frac{d}{d{q}_1}\left[\frac{1}{2}\frac{k_{\text{e}}{q}_1^2}{R_1}+\frac{k_{\text{e}}}{2}\frac{{\left(Q-q_1\right)}^2}{R_2}\right]=0\\ \frac{k_{\text{e}}{q}_1}{R_1}+\frac{k_{\text{e}}\left(Q-q_1\right)}{R_2}\left(-1\right)=0\\ q_1=\frac{R_{1}Q}{R_1+R_2}\end{array}$

Conclusion:

Therefore, the value of $q_1$ is $q_1=\frac{R_{1}Q}{R_1+R_2}$.

(d)

To determine

The value of $q_2$ from part (c).

Answer to Problem 40P

The value of $q_2$ is $q_2=\frac{R_{2}Q}{R_1+R_2}$.

Explanation of Solution

Given info: The radii of two conducting sphere is $R_1$ and $R_2$. The total charge shared between them is $Q$. The charge on first and second sphere is $q_1$ and $q_2$ respectively.

The value of $q_1$ is,

$q_1=\frac{R_{1}Q}{R_1+R_2}$.

The sum of charge of both sphere are,

$\begin{array}{c}Q=q_1+q_2\\ q_2=Q-q_1\end{array}$

Substitute $\frac{R_{1}Q}{R_1+R_2}$ for $q_2$ in above equation.

$\begin{array}{c}{q}_2=Q-\frac{R_{1}Q}{R_1+R_2}\\ =\frac{R_{2}Q}{R_1+R_2}\end{array}$

Conclusion:

Therefore, the value of $q_2$ is $q_2=\frac{R_{2}Q}{R_1+R_2}$.

(e)

To determine

The potential of each sphere.

Answer to Problem 40P

The potential of each sphere is $\frac{k_{\text{e}}Q}{R_1+R_2}$.

Explanation of Solution

Given info: The radii of two conducting sphere is $R_1$ and $R_2$. The total charge shared between them is $Q$. The charge on first and second sphere is $q_1$ and $q_2$ respectively.

Write the expression to calculate the potential of first sphere.

$V_1=\frac{k_{\text{e}}{q}_1}{R_1}$

Substitute $\frac{R_{1}Q}{R_1+R_2}$ for $q_1$ in above equation.

$\begin{array}{c}{V}_1=\frac{k_{\text{e}}}{R_1}\times \frac{R_{1}Q}{R_1+R_2}\\ =\frac{k_{\text{e}}Q}{R_1+R_2}\end{array}$

Write the expression to calculate the potential of second sphere.

$V_2=\frac{k_{\text{e}}{q}_2}{R_2}$

Substitute $\frac{R_{2}Q}{R_1+R_2}$ for $q_1$ in above equation.

$\begin{array}{c}{V}_2=\frac{k_{\text{e}}}{R_2}\times \frac{R_{2}Q}{R_1+R_2}\\ =\frac{k_{\text{e}}Q}{R_1+R_2}\end{array}$

Thus, the potential of each sphere is $\frac{k_{\text{e}}Q}{R_1+R_2}$.

Conclusion:

Therefore, the potential of each sphere is $\frac{k_{\text{e}}Q}{R_1+R_2}$.

(f)

To determine

The potential difference between the spheres.

Answer to Problem 40P

The potential difference between the spheres is zero.

Explanation of Solution

Given info: The radii of two conducting sphere is $R_1$ and $R_2$. The total charge shared between them is $Q$. The charge on first and second sphere is $q_1$ and $q_2$ respectively.

The potential difference is,

$\Delta V=V_1-V_2$

Substitute $\frac{k_{\text{e}}Q}{R_1+R_2}$ $V_1\text{and}{V}_2$ in above equation.

$\begin{array}{c}\Delta V=\frac{k_{\text{e}}Q}{R_1+R_2}-\frac{k_{\text{e}}Q}{R_1+R_2}\\ =0\end{array}$

Conclusion:

Therefore, the potential difference between the spheres is zero.