# Energy of a capacitor and an electric field

Elementary work of external forces to move charge dq in electric field of a capacitor

$dA=dq*\left({\phi }_{1}–{\phi }_{2}\right)=dq\frac{q}{C}$

Total work is

$A={\int }_{0}^{Q}\frac{dqq}{C}=\frac{{Q}^{2}}{2C}$

this work determines total energy stored in a capacitor, Q is a total capacitor charge.

$Q=C\left({\phi }_{1}–{\phi }_{2}\right)$

and energy of a charged capacitor

$W=\frac{C\left({\phi }_{1}–{\phi }_{2}{\right)}^{2}}{2}$

Let’s express these characteristics through the electric field parameters. For flat capacitors

${\phi }_{1}–{\phi }_{1}=Ed$

and

$C=\frac{\epsilon {\epsilon }_{0}S}{d}$

then

V – is is the volume between capacitor plates. Therefore, volume energy density can be found for a capacitor

${w}_{e}=\frac{W}{V}=\frac{\epsilon {\epsilon }_{0}{E}^{2}}{2}$

Volume energy density has local characteristics, and it corresponds to the piece of a capacitor where the electric field is uniform and equal to E. Let’s consider the term of volume energy density, on the example of non-uniform electric field. Take a piece of space with volume dV, that characterises radius-vector r. Volume density of energy in general is the value expressed by the formula

${w}_{e}=\frac{d{W}_{e}}{dV}$

Figure 22 dWis the energy of the small piece of electric field. So if we know the electric field E(r), we can calculate the energy of any piece of field with finite dimensions Ω. Therefore,

${w}_{e}=\frac{\epsilon {\epsilon }_{0}{E}^{2}\left(\overline{r}\right)}{2,{W}_{e}}={\int }_{\Omega }{w}_{e}dV={\int }_{\Omega }\frac{\epsilon {\epsilon }_{0}{E}^{2}\left(\overline{r}\right)}{2dV}$

These formulas work for the uniform capacitor material with permittivity ε=const.

Electric field in dielectrics. Electric dipole
Dipoles play an important role in describing dielectrics in the electric field. Electric dipole is a system of two equal charges q, with opposite signs, placed on the distance l. A Dipole moment is the main dipole characteristic.

from negative to positive charge – called the dipole shoulder. Figure 23. For example, molecules

${H}_{2}O,HCl,N{H}_{3}$

are dipoles. Let’s find out how dipoles behave in different kinds of electric fields.

Uniform field
Electric field is constant in any point of space, forces affecting the charges +q and –q, are equal with an opposite sign. Resulting force is 0. The dipole of these forces is not 0, if the dipole is not oriented parallel to the electric field lines. Figure 24. Force momentum, according to the axis normal to the figure plane: N=F*l*sinα=q*E*l*sinα=p*E*sinα,

$\overline{N}=\left[\overline{p}*\overline{E}\right]$

This means a uniform field turns the dipole along the field lines.

Non-uniform field
In non-uniform fields, the dipole is affected by turning the moment and electric force. Assuming that field changes only in one direction, Figure 25. Projection of force on the axis x is

${F}_{x}={F}_{x}^{–}+{F}_{x}^{+}=q{E}_{x}^{–}+q{E}_{x}^{+}=q*dE=q*\partial E/\partial xdx,dx=l*cos\alpha$

Then

${F}_{x}=q*\frac{\partial E}{\partial x}*l*cos\alpha =q*l*\frac{\partial E}{\partial x}*cos\alpha =p*\frac{\partial E}{\partial x}*cos\alpha$

Dipoles are oriented along the field. Fhas a positive or negative sign depending on the electric field, and if it is increasing or decreasing along the 0X axis. Dipoles are drawn into the stronger part of field. Electric field increment is

$\overline{dE}=\left(\frac{\partial E}{\partial x}dx\right)\overline{{e}_{x}}+\left(\frac{\partial E}{\partial y}dy\right)\overline{{e}_{y}}+\left(\frac{\partial E}{\partial z}dz\right)\overline{{e}_{z}}$

then the force affecting the dipole is

$\overline{F}=\left(\frac{\partial E}{\partial x}{p}_{x}\right)\overline{{e}_{x}}+\left(\frac{\partial E}{\partial y}{p}_{y}\right)\overline{{e}_{y}}+\left(\frac{\partial E}{\partial z}{p}_{z}\right)\overline{{e}_{z}}$

The conclusion is that dipoles are orienting along the electric field lines and drawing into the electric field with a bigger intensity.

Dipole energy in the electric field

$W=q{\phi }^{+}–q{\phi }^{–}.{\phi }^{+}–{\phi }^{–}=\frac{\partial \phi }{\partial x}dx=\frac{\partial \phi }{\partial x}l*cos\alpha ,{E}_{x}=–grad\phi$

then

Force of electric field

$\overline{F}=–gradW,{F}_{x}=–\frac{\partial W}{\partial x}=\frac{p\partial E}{\partial x}cos\alpha$
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