Electrostatic potential and charges on conductors that are closed to each other can be put in relation with the capacitance matrix .

Can the energy of the system of two (or more) conductors be rewritten as the sum of a part due to each conductor and another one that is due to a "shared" energy of the two conductors?

Consider two conductors of capacitance ,charges and potentials $q_1$, $C_1$, $V_1$, $q_2$, $C_2$, $V_2$.

The energy of the system is by definition $$U=q_1 V_1+q_2 V_2$$

The matrix of capacitance is the 2x2 symmetric matrix such that

$$\begin{pmatrix} q_1 \\ q_2 \end{pmatrix}=\begin{pmatrix}c_{11} & c_{12}\\ c_{12}& c_{21} \end{pmatrix}\begin{pmatrix}V_1 \\ V_2\end{pmatrix}$$

Can I express $U$ as something like the following?


Where $...$ stays for an expression that includes the charges, the potentials and the coefficients $c_{11},c_{12},c_{23}$. This expression should represent the "shared" energy of the two conductors.

Example (which I wonder how to generalize)

Two conductiong spheres have the parameters indicated above and are at a big distance $x$ (induction influence is neglected). The energy of the system can be written as

$$U=\frac{q_1^2}{2C_1}+\frac{q_2^2}{2C_2}+\frac{q_1 q_2}{4 \pi \epsilon_0 x}$$

In this case the expression I'm looking for is $\frac{q_1 q_2}{4 \pi \epsilon_0 x}$, but how can one in general write this term (if it is possible to do it)?


1 Answer 1


If you assume that $$U=\frac {1}{2}q_1V_1+\frac {1}{2}q_2V_2$$ is correct for the total energy, you simply solve the matrix capacitance equation for $q_1$ and $q_2$, insert them into the system energy equation, and order the terms according to the products of $V_1, V_2$ Thus you'll obtain for the energy $$U=\frac {1}{2}( c_{11}V_1^2+2c_{12}V_1·V_2+C_{22}V_2^2)$$ where symmetry of the matrix $c_{12}=c_{21}$ is used.

In the general case of n conductors, you can always diagonalize the symmetric capacitance matrix by a suitable orthogonal transformation of the "coordinates" $V_1,V_2,...$ to coordinates $V_1^*, V_2^*,...$ to obtain a quadratic form for the energy $U=\frac {1}{2}\sum_{i=1}^n c_iV_i^*{^2}$.

  • $\begingroup$ Thanks for the answer! Usually I see that the energy associated to the single conductor is written as $ \frac{1}{2} C_1 V_1^2$ and not as $c_{11} V_1^2$, as in your answer. Is this the same quantity? Is maybe the difference due to the fact that, in general $c_{11} \neq C_1$? $\endgroup$
    – Sørën
    Nov 15, 2016 at 17:15
  • $\begingroup$ Edit: I see now that the expression of $U$ in the question was wrong: it is not $q_1V_1+q_2 V_2$ but $\frac{1}{2} q_1 V_1+\frac{1}{2} q_2 V_2$. So can I say that $$U= \frac{1}{2}c_{11}V_1^2+2c_{12}V_1·V_2+\frac{1}{2}c_{22}V_2^2$$? $\endgroup$
    – Sørën
    Nov 15, 2016 at 17:42
  • 1
    $\begingroup$ @Soren - I suspected that. The usual formula for the potential energy of a system of conductors is $$U=(1/2)\sum_{i=1}^{n} q_i V_i$$ thus, for the considered case, the correct formula is the one given divided by 2: $$U= \frac{1}{2} c_{11}V_1^2+c_{12}V_1·V_2+\frac{1}{2} C_{22}V_2^2$$ $\endgroup$
    – freecharly
    Nov 15, 2016 at 18:28
  • 1
    $\begingroup$ @Soren - I also corrected the answer by the factor 1/2. The second term has no factor 1/2 because it appears twice due to the symmetry. $\endgroup$
    – freecharly
    Nov 15, 2016 at 18:36

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