# Energy stored in an electrostatic system

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1. Aug 11, 2017

### Pushoam

<Moderator's note: Thread moved from a technical forum, so homework template missing>

To do: To find an expression for energy stored in an electrostatic system with charge density $\rho$ and volume R.
[ I am using R to denote the region filled with the given charge density as I want to keep 'V' reserved for potential ].

By the term "energy of the system " what we mean is : work required to assemble the system

So, I have to calculate work required to assemble the system i.e. W.

Let's consider the system at the time when its volume is Δ R and I bring charge dq to a point M with position vector $\vec r$ on its surface.

Work done by me in this process is
dW = V($\vec r$) dq

$V \left ( \vec r \right ) = k \int _{ΔR}~ \frac { \rho (\vec {r'}) }{r"} d \, τ'$
$dq = \rho (\vec {r}) dτ$
So, $dW = k \int _R~ \frac { \rho (\vec {r'}) }{r"} d \, τ' ~ \rho (\vec {r}) dτ$
$W = \int _R~ \rho (\vec {r}) \{ k \int _{ΔR}~ \frac { \rho (\vec {r'}) }{r"} d \, τ'\} d\,τ$

where ΔR depends on r.

Is this correct so far?

Another way:

W = ½ $\int_R ~ \rho (\vec r) V (\vec r) d \, τ$
$\rho (\vec r) V (\vec r) = ε_0 (∇⋅ \vec E ) V = ε_0 ( ∇⋅( V \vec E ) - \vec E⋅ ∇ V )$
$W = ½ \int_R~ ε_0 ( ∇⋅ (V \vec E ) - \vec E⋅ ∇ V ) d \,τ$
$= ½ \int_S ~ ε_0 V \vec E ⋅ d \vec a + ½ \int_R ε_0 E^2 d \, τ$
Taking the region of integration to be infinite, (as the charge density outside R is 0), the first integral on R.H.S. becomes 0. Hence, we have,
$W = \frac {\epsilon _0 } 2 \int_{all space} E^2 d \, \tau$

In case of dielectrics, this will be the work required for assembling both free charges and bound charges.

For dielectrics, $\rho = \rho_b + \rho_f,$
So,
W = ½ $\int_R ~ \rho (\vec r) V (\vec r) d \, τ$
= ½ $\int_R~ (\rho_b + \rho_f )V (\vec r) d \, τ$
W =½ $\int_R~ \rho_b V d \, τ + ½ \int_R \rho_f V (\vec r) d \, τ$

Can I take the first term on the R.H.S. as the work required to assemble the bound charges and the second term as the work required to assemble the free charges ?

If yes, then,

$W_f = ½ \int_R \rho_f V (\vec r) d \, τ \\ \rho_f=∇.\vec D$
And doing some further calculation,
$W_f = ½ \int_{all space} \vec D . \vec E ~d \, τ$

Is this correct so far?

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2. Aug 12, 2017

### Pushoam

Calculating work done required in building free charges up from zero to the final configuration in dielectrics

Here, the volume of the system( dielectrics ) R is fixed. So, as I bring free charge Δq to the system, its free charge density $\rho_f$ gets changed by $Δ \rho_f$. Change in free charge density creates change in bound charge density ( only $\rho_b$ or both$\rho_b$ and $σ_b$ ?).

$Δq = \int_R (ρ_f + Δ ρ_f ) d \, τ - \int_R ρ_f d \, τ = \int_R Δρ_f d \, τ$

Work done on bringing this incremental charge from infinity to a point with position vector $\vec r$ inside the dielectrics is

$ΔW = Δq V(\vec r )$ (1)

$ΔW = V(\vec r ) \int_R Δρ_f d \, τ$ (1.1)

But, the Griffith says in eq. 4.56 3rd ed. : $ΔW = \int_R Δρ_f V(\vec r ) d \, τ$
So, something may be wrong in (1).
I think Δq gets spread in all over the dielectrics. It doesn't remain at $\vec r$ in the dielectrics.
So, eq (1) has to be modified.
Let's say that a very small amount of Δq i.e. Δ (Δq) remains at $\vec r$ after the spreading. Now, work done on this charge is

$Δ(ΔW) = Δ(Δq) V( \vec r )$

On integrating both sides,

$ΔW = Δ\int _R Δ(Δρ_f) V(\vec r ) d \, τ$

This also doesn't match the Griffith's expression.
I guess the way Griffith's brings the free charge is following:
He doesn't bring the charge Δq at $\vec r$. He brings the charge Δq over all the region of dielectrics simultaneously so that its free charge density gets increased by $Δρ_f$. Now, $Δ ρ_f (\vec r ) V(\vec r ) d \, τ$ is the work done in bringing the charge $Δ ρ_f (\vec r ) d \, τ$ at $\vec r$. Hence, the work done in bringing Δq all over the dielectrics is

$ΔW = \int _R~ Δ ρ_f (\vec r ) ~ V(\vec r ) d \, τ$

Where $V (\vec r )$ is potential due to the system at $\vec r$ at the time when the charge $Δ ρ_f (\vec r ) d \, τ$ is brought to $\vec r$.

$Δρ_f = ∇ρ_f ⋅ d \vec r = ∇ (∇⋅ \vec D )⋅ d \vec r$

Now, Griffiths takes $Δ ρ_f = ∇. (Δ \vec D )$.
This means,
\begin {align} Δρ_f & = Δ (∇. \vec D ) & = ∇. (Δ \vec D ) \\& ⇒ [ ∇(∇. \vec D)]. d \vec r & = ∇. ( ∇ D_j .d \vec r ) \hat x_j \\ & ⇒ \frac { \partial { [ \frac {\partial D_j } {\partial x_j } ] } } {\partial {x_i } } d x_i & = \frac { \partial { [ \frac {\partial D_j } {\partial x_k } d x_k ] } } {\partial {x_j } } \end {align}

How to prove the equality in the last equation?

Since I understand the following steps, I am copying it from Griffiths.

Now, since we have to calculate W instead of ΔW, what Griffith does is : $(Δ \vec D ) . \vec E = \frac {Δ(\vec D .\vec E )} 2$

This I can show as following:

$(Δ \vec D ) . \vec E = E_i (∇ D_i . d\vec r ) = \frac { Δ (E_i D_i )} 2 = \frac {Δ(\vec D .\vec E )} 2$

while taking the integration over all space, it is assumed that space outside R is vacuum.
Is this assumption necessary?

Now, Griffith takes Δ outside the integration. Can anyone please explain me how can one show the correctness of this step?

Can anyone please explain me the underlined part in the following footnote?

I am attaching Griffith's derivation of eqn. 2.42 and 4.58 given in 3rd ed.