Chapter 26, Problem 26.40P

Consider two conducting spheres with radii R₁ and R₂ separated by a distance much greater than cither radius. A total charge Q is shared between the spheres. We wish to show that when the electric potential energy of the system has a minimum value, the potential difference between the spheres is zero. The total charge Q is equal to q₁ + q₂, where q₁ represents the charge on the first sphere and q₂ the charge on the second. Because the spheres are very far apart, you can assume the charge of each is uniformly distributed over its surface. (a) Show that the energy associated with a single conducting sphere of radius R and charge q surrounded by a vacuum is U_E = k_eq²/2R. (b) Find the total energy’ of the system of two spheres in terms of the total charge Q, and the radii and R₁ and R₂. (c) To minimize the energy, differentiate the result to part (b) with respect to q₁ and set the derivative equal to zero. Solve for q₁ in terms of Q and the radii. (d) From the result to part (c), find the charge q₂. (e) Find the potential of each sphere. (f) What is the potential difference between the spheres?

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 26.40P

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 $U_{\text{E}}=\frac{k_{\text{e}}{q}^2}{2R}$.

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 26.40P

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

To determine

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

Answer to Problem 26.40P

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

To determine

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

Answer to Problem 26.40P

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 $q_2=\frac{R_{2}Q}{R_1+R_2}$.

To determine

The potential of each sphere.

Answer to Problem 26.40P

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 $\frac{k_{\text{e}}Q}{R_1+R_2}$.

To determine

The potential difference between the spheres.

Answer to Problem 26.40P

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.