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Looking upon a certain solution of a problem in Griffiths regarding the work done in arranging two concentric spheres with inner shell having charge $q$ and outer shell with charge $-q$ and I found a statement as superposition of work, and an expression as $$W_{\text{total}}=W_{1}+W_2+\epsilon_{o}\int (E_1\cdot E_2 )d\tau$$

Is there any proof for this expression, as I'm not able to find any references for this expression.

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  • $\begingroup$ What are $W_1$ and $W_2$? $\endgroup$
    – probably_someone
    Commented Dec 16, 2018 at 13:10
  • $\begingroup$ $W_{1}=\frac{q^2}{8 \pi \epsilon_{o} a}$ and $W_{2}=\frac{q^2}{8 \pi \epsilon_{o} b}$ , where $a$ is the radius of inner sphere and $b$ is the radius of outer sphere. $\endgroup$
    – Sahil Silare
    Commented Dec 16, 2018 at 13:13

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If the electric fields due to the two charges are $\vec E_1(r)$ and $\vec E_2(r)$ then the resultant field due to the two charges is $E(r) = \vec E_1(r)+\vec E_2(r)$

The energy density is $\frac 12\epsilon_0 \vec E(r)^2 = \frac 12\epsilon_0(\vec E_1(r)+\vec E_2(r))^2= \frac 12\epsilon_0(E_1^2(r) + E_2^2(r) + 2 \vec E_1(r)\cdot \vec E_2(r))$

and on integration over a volume the result follows noting that in your example the third term is nagative.

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The total work required could be split into three parts: the work to bring the first shell together (if the second shell were not there). the work to bring the second shell together (if the first were not there) and the work to bring the second shell near the first. Those are the three terms in the equation you give.

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  • $\begingroup$ How the third term $\epsilon_{o} \int E_1 \cdot E_2 d\tau$ came? $\endgroup$
    – Sahil Silare
    Commented Dec 16, 2018 at 13:28
  • $\begingroup$ The third term is the sum of potential energies between each charge in the inner shell and each charge in the outer shell. Treating the charge as continuous, the sum becomes an integral. $\endgroup$
    – Rich006
    Commented Dec 17, 2018 at 22:28

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