Electric Potential Energy - Energy Stored in An Electrostatic Field Distribution | Energy Stored Electrostatic Field Distribution

Energy Stored in An Electrostatic Field Distribution

The energy density, or energy per unit volume, of the electrostatic field of a continuous charge distribution is:

Outline of proof
One may take the equation for the electrostatic potential energy of a continuous charge distribution and put it in terms of the electrostatic field. Since Gauss' law for electrostatic field in differential form states where is the electric field vector is the total charge density including dipole charges bound in a material is the permittivity of free space, then, $\begin{align} U & = \frac{1}{2}\int \limits_{\text{all space}} \rho(r) \Phi(r) \, dV \\ & = \frac{1}{2}\int \limits_{\text{all space}} \varepsilon_0(\mathbf{\nabla}\cdot{\mathbf{E}})\Phi \, dV \end{align}$ so, now using the following divergence vector identity we have using the divergence theorem and taking the area to be at infinity where $\begin{align} U & = \overbrace{\frac{\varepsilon_0}{2}\int\limits_{{}^\text{boundary}_\text{ of space}} \Phi\mathbf{E}\cdot d\mathbf A}^{0} - \frac{\varepsilon_0}{2}\int \limits_{\text{all space}} (-\mathbf{E})\cdot\mathbf{E} \, dV \\ & = \int \limits_{\text{all space}} \frac{1}{2}\varepsilon_0\left\|{\mathbf{E}}\right\|^2 \, dV. \end{align}$ So, the energy density, or energy per unit volume of the electrostatic field is:

Outline of proof

One may take the equation for the electrostatic potential energy of a continuous charge distribution and put it in terms of the electrostatic field.

Since Gauss' law for electrostatic field in differential form states

where

is the electric field vector
is the total charge density including dipole charges bound in a material
is the permittivity of free space,

then,

$\begin{align} U & = \frac{1}{2}\int \limits_{\text{all space}} \rho(r) \Phi(r) \, dV \\ & = \frac{1}{2}\int \limits_{\text{all space}} \varepsilon_0(\mathbf{\nabla}\cdot{\mathbf{E}})\Phi \, dV \end{align}$

so, now using the following divergence vector identity

we have

using the divergence theorem and taking the area to be at infinity where

$\begin{align} U & = \overbrace{\frac{\varepsilon_0}{2}\int\limits_{{}^\text{boundary}_\text{ of space}} \Phi\mathbf{E}\cdot d\mathbf A}^{0} - \frac{\varepsilon_0}{2}\int \limits_{\text{all space}} (-\mathbf{E})\cdot\mathbf{E} \, dV \\ & = \int \limits_{\text{all space}} \frac{1}{2}\varepsilon_0\left|{\mathbf{E}}\right|^2 \, dV. \end{align}$

So, the energy density, or energy per unit volume of the electrostatic field is:

Read more about this topic: Electric Potential Energy

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