Question

Two identical capacitors each of capacity C with plate separation $d$ are placed $r$ distance apart ($r>>d$) as shown in diagram. Mutual positions are as shown in diagram. Mutual potential energy of two capacitors is (each capacitor is charged to potential of $V$ each)

Answer 1

-\frac{C^2V^2d^2}{2\pi\epsilon_0r^3}

Answer 2

The mutual potential energy between two charged capacitors can be calculated by considering the interaction energy between all pairs of charges on the plates of the two capacitors.

Let the left capacitor have charges $+Q$ on the left plate and $-Q$ on the right plate. Let the right capacitor have charges $+Q$ on the left plate and $-Q$ on the right plate. The distance between the plates of each capacitor is $d$. The distance between the nearest plates (positive of left and positive of right) is $r$. The distance between the positive plate of the left capacitor and the negative plate of the right capacitor is $r+d$. The distance between the negative plate of the left capacitor and the positive plate of the right capacitor is $r+d$. The distance between the negative plate of the left capacitor and the negative plate of the right capacitor is $r$.

Let $q_{L+} = +Q$, $q_{L-} = -Q$, $q_{R+} = +Q$, $q_{R-} = -Q$. The distances between the charges are: $R_{L+R+} = r$ $R_{L+R-} = r+d$ $R_{L-R+} = r+d$ $R_{L-R-} = r$

The mutual potential energy is the sum of the potential energies of interaction between these pairs of charges: $U_{mutual} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_{L+} q_{R+}}{R_{L+R+}} + \frac{q_{L+} q_{R-}}{R_{L+R-}} + \frac{q_{L-} q_{R+}}{R_{L-R+}} + \frac{q_{L-} q_{R-}}{R_{L-R-}} \right)$ $U_{mutual} = \frac{1}{4\pi\epsilon_0} \left( \frac{(+Q)(+Q)}{r} + \frac{(+Q)(-Q)}{r+d} + \frac{(-Q)(+Q)}{r+d} + \frac{(-Q)(-Q)}{r} \right)$ $U_{mutual} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q^2}{r} - \frac{Q^2}{r+d} - \frac{Q^2}{r+d} + \frac{Q^2}{r} \right)$ $U_{mutual} = \frac{Q^2}{4\pi\epsilon_0} \left( \frac{2}{r} - \frac{2}{r+d} \right)$ $U_{mutual} = \frac{2Q^2}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{r+d} \right)$ $U_{mutual} = \frac{2Q^2}{4\pi\epsilon_0} \left( \frac{r+d - r}{r(r+d)} \right) = \frac{2Q^2}{4\pi\epsilon_0} \frac{d}{r(r+d)}$

We are given that each capacitor is charged to a potential $V$, so $Q = CV$. $U_{mutual} = \frac{2(CV)^2}{4\pi\epsilon_0} \frac{d}{r(r+d)} = \frac{2C^2 V^2 d}{4\pi\epsilon_0 r(r+d)}$

Since $r \gg d$, we can approximate $r+d \approx r$. $U_{mutual} \approx \frac{2C^2 V^2 d}{4\pi\epsilon_0 r \cdot r} = \frac{2C^2 V^2 d}{4\pi\epsilon_0 r^2}$

Let's reconsider the distances based on the diagram. The diagram shows the distance $r$ between the centers of the capacitors along the horizontal direction. Let the left capacitor be centered at $x=0$, and the right capacitor be centered at $x=r$. The left capacitor has plates at $x=-d/2$ (positive) and $x=d/2$ (negative). The right capacitor has plates at $x=r-d/2$ (positive) and $x=r+d/2$ (negative). Charges: $q_{L+} = +Q$ at $x_{L+} = -d/2$, $q_{L-} = -Q$ at $x_{L-} = d/2$. $q_{R+} = +Q$ at $x_{R+} = r-d/2$, $q_{R-} = -Q$ at $x_{R-} = r+d/2$.

Distances between pairs of charges from different capacitors: $R_{L+R+} = |x_{R+} - x_{L+}| = |r-d/2 - (-d/2)| = |r-d/2+d/2| = r$ $R_{L+R-} = |x_{R-} - x_{L+}| = |r+d/2 - (-d/2)| = |r+d/2+d/2| = |r+d| = r+d$ $R_{L-R+} = |x_{R+} - x_{L-}| = |r-d/2 - d/2| = |r-d| = r-d$ (since $r \gg d$, $r-d > 0$) $R_{L-R-} = |x_{R-} - x_{L-}| = |r+d/2 - d/2| = |r| = r$

Mutual potential energy: $U_{mutual} = \frac{1}{4\pi\epsilon_0} \left( \frac{(+Q)(+Q)}{r} + \frac{(+Q)(-Q)}{r+d} + \frac{(-Q)(+Q)}{r-d} + \frac{(-Q)(-Q)}{r} \right)$ $U_{mutual} = \frac{Q^2}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{r+d} - \frac{1}{r-d} + \frac{1}{r} \right)$ $U_{mutual} = \frac{Q^2}{4\pi\epsilon_0} \left( \frac{2}{r} - \left( \frac{1}{r+d} + \frac{1}{r-d} \right) \right)$ $U_{mutual} = \frac{Q^2}{4\pi\epsilon_0} \left( \frac{2}{r} - \frac{(r-d) + (r+d)}{(r+d)(r-d)} \right) = \frac{Q^2}{4\pi\epsilon_0} \left( \frac{2}{r} - \frac{2r}{r^2 - d^2} \right)$ $U_{mutual} = \frac{2Q^2}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{r}{r^2 - d^2} \right) = \frac{2Q^2}{4\pi\epsilon_0} \left( \frac{r^2 - d^2 - r^2}{r(r^2 - d^2)} \right) = \frac{2Q^2}{4\pi\epsilon_0} \left( \frac{-d^2}{r(r^2 - d^2)} \right)$ $U_{mutual} = - \frac{2Q^2 d^2}{4\pi\epsilon_0 r(r^2 - d^2)}$

Since $r \gg d$, $r^2 - d^2 \approx r^2$. $U_{mutual} \approx - \frac{2Q^2 d^2}{4\pi\epsilon_0 r \cdot r^2} = - \frac{2Q^2 d^2}{4\pi\epsilon_0 r^3}$

Substitute $Q = CV$: $U_{mutual} \approx - \frac{2(CV)^2 d^2}{4\pi\epsilon_0 r^3} = - \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r^3}$

Let's use the dipole approximation. Each capacitor is a dipole with moment $p = Qd$. For the left capacitor, the positive plate is at $x=-d/2$ and the negative plate is at $x=d/2$. The dipole moment points from negative to positive, so it points from $x=d/2$ to $x=-d/2$, i.e., in the $-\hat{i}$ direction. $\vec{p}_L = -Qd \hat{i}$. For the right capacitor, the positive plate is at $x=r-d/2$ and the negative plate is at $x=r+d/2$. The dipole moment points from negative to positive, so it points from $x=r+d/2$ to $x=r-d/2$, i.e., in the $-\hat{i}$ direction. $\vec{p}_R = -Qd \hat{i}$. The vector from the center of the left capacitor ($x=0$) to the center of the right capacitor ($x=r$) is $\vec{r} = r \hat{i}$. The potential energy of interaction between two dipoles is $U = \frac{1}{4\pi\epsilon_0 r^3} [\vec{p}_1 \cdot \vec{p}_2 - 3 (\vec{p}_1 \cdot \hat{r}) (\vec{p}_2 \cdot \hat{r})]$. $\vec{p}_L \cdot \vec{p}_R = (-Qd \hat{i}) \cdot (-Qd \hat{i}) = Q^2 d^2$. $\vec{p}_L \cdot \hat{r} = (-Qd \hat{i}) \cdot \hat{i} = -Qd$. $\vec{p}_R \cdot \hat{r} = (-Qd \hat{i}) \cdot \hat{i} = -Qd$. $U_{mutual} = \frac{1}{4\pi\epsilon_0 r^3} [Q^2 d^2 - 3 (-Qd)(-Qd)] = \frac{1}{4\pi\epsilon_0 r^3} [Q^2 d^2 - 3Q^2 d^2] = \frac{-2Q^2 d^2}{4\pi\epsilon_0 r^3}$. Substituting $Q = CV$, $U_{mutual} = - \frac{2(CV)^2 d^2}{4\pi\epsilon_0 r^3} = - \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r^3}$.

Let's consider the case where the positive plate of the left capacitor faces the negative plate of the right capacitor. Left capacitor: $+Q$ on left, $-Q$ on right. Right capacitor: $-Q$ on left, $+Q$ on right. Distance between centers is $r$. Charges: $q_{L+} = +Q$ at $x=-d/2$, $q_{L-} = -Q$ at $x=d/2$. $q_{R-} = -Q$ at $x=r-d/2$, $q_{R+} = +Q$ at $x=r+d/2$. Distances: $R_{L+R-} = |r-d/2 - (-d/2)| = r$ $R_{L+R+} = |r+d/2 - (-d/2)| = r+d$ $R_{L-R-} = |r-d/2 - d/2| = r-d$ $R_{L-R+} = |r+d/2 - d/2| = r$ $U_{mutual} = \frac{1}{4\pi\epsilon_0} \left( \frac{(+Q)(-Q)}{r} + \frac{(+Q)(+Q)}{r+d} + \frac{(-Q)(-Q)}{r-d} + \frac{(-Q)(+Q)}{r} \right)$ $U_{mutual} = \frac{Q^2}{4\pi\epsilon_0} \left( -\frac{1}{r} + \frac{1}{r+d} + \frac{1}{r-d} - \frac{1}{r} \right) = \frac{Q^2}{4\pi\epsilon_0} \left( -\frac{2}{r} + \frac{2r}{r^2 - d^2} \right)$ $U_{mutual} = \frac{2Q^2}{4\pi\epsilon_0} \left( -\frac{1}{r} + \frac{r}{r^2 - d^2} \right) = \frac{2Q^2}{4\pi\epsilon_0} \left( \frac{-r^2 + d^2 + r^2}{r(r^2 - d^2)} \right) = \frac{2Q^2}{4\pi\epsilon_0} \frac{d^2}{r(r^2 - d^2)}$ For $r \gg d$, $U_{mutual} \approx \frac{2Q^2 d^2}{4\pi\epsilon_0 r^3}$. Substituting $Q = CV$, $U_{mutual} \approx \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r^3}$.

The problem statement and diagram show the first case where positive faces positive and negative faces negative. So the first calculation is correct. $U_{mutual} = - \frac{2Q^2 d^2}{4\pi\epsilon_0 r(r^2 - d^2)}$. Using the approximation $r \gg d$, $U_{mutual} \approx - \frac{2Q^2 d^2}{4\pi\epsilon_0 r^3}$. Substitute $Q = CV$. $U_{mutual} \approx - \frac{2(CV)^2 d^2}{4\pi\epsilon_0 r^3} = - \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r^3}$.

Let's express $\frac{1}{4\pi\epsilon_0}$ in terms of $C$ and $d$. For a parallel plate capacitor with area $A$, $C = \frac{\epsilon_0 A}{d}$, so $\epsilon_0 = \frac{Cd}{A}$. $\frac{1}{4\pi\epsilon_0} = \frac{A}{4\pi Cd}$. This introduces the area $A$, which is not given.

Let's use the dipole moment $p = Qd = (CV)d$. $U_{mutual} \approx - \frac{2 p^2}{4\pi\epsilon_0 r^3} = - \frac{2 (CVd)^2}{4\pi\epsilon_0 r^3}$. We need to find a relationship between $C$, $d$, and $\epsilon_0$ that doesn't involve the area $A$. This is not possible for a general parallel plate capacitor unless some other information is given or implied.

Let's re-examine the exact expression: $U_{mutual} = \frac{2Q^2}{4\pi\epsilon_0} \frac{d}{r(r+d)}$ for the first configuration (positive facing negative). $U_{mutual} = \frac{2C^2 V^2 d}{4\pi\epsilon_0 r(r+d)}$.

The second exact expression: $U_{mutual} = - \frac{2Q^2 d^2}{4\pi\epsilon_0 r(r^2 - d^2)}$ for positive facing positive. $U_{mutual} = - \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r(r^2 - d^2)}$.

Given the diagram, the second case is the correct one. The mutual potential energy is negative, which indicates an attractive force between the two capacitors. Let's check the forces. The positive plate of the left capacitor attracts the negative plate of the right capacitor, and repels the positive plate of the right capacitor. The negative plate of the left capacitor attracts the positive plate of the right capacitor, and repels the negative plate of the right capacitor. Forces: Attraction between $L_+$ and $R_-$ is proportional to $1/(r+d)^2$. Repulsion between $L_+$ and $R_+$ is proportional to $1/r^2$. Attraction between $L_-$ and $R_+$ is proportional to $1/(r-d)^2$. Repulsion between $L_-$ and $R_-$ is proportional to $1/r^2$. Since $r-d < r < r+d$, the attractive force between $L_-$ and $R_+$ is the strongest. The repulsive forces are equal. The attractive force between $L_+$ and $R_-$ is the weakest. The net force is attractive. This is consistent with a negative potential energy.

The final answer should be the exact expression or the leading term in the approximation. Since $r \gg d$, the approximation is usually expected.

The mutual potential energy is $U_{mutual} = - \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r(r^2 - d^2)}$. Using the approximation $r \gg d$, $U_{mutual} \approx - \frac{2C^2 V^2 d^2}{4\pi\epsilon_0 r^3}$.

Let's write the answer in terms of $C, V, d, r, \epsilon_0$.

The final answer is $-\frac{C^2V^2d^2}{2\pi\epsilon_0r^3}$.