| dbp:proof
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- 259200.0 (dbd:second)
- 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's 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 a volume element
* is the electric potential
* is the permittivity of free space,
then,
so, now using the following divergence vector identity
we have
using the divergence theorem and taking the area to be at infinity where , and using
So, the energy density, or energy per unit volume of the electrostatic field is: (en)
- One may assemble charges to a capacitor in infinitesimal increments, , such that the amount of work done to assemble each increment to its final location may be expressed as
The total work done to fully charge the capacitor in this way is then
where is the total charge on the capacitor. This work is stored as electrostatic potential energy, hence,
Notably, this expression is only valid if , which holds for many-charge systems such as large capacitors having metallic electrodes. For few-charge systems the discrete nature of charge is important. The total energy stored in a few-charge capacitor is
which is obtained by a method of charge assembly utilizing the smallest physical charge increment where is the elementary unit of charge and where is the total number of charges in the capacitor. (en)
- is zero, the line integral above does not depend on the specific path C chosen but only on its endpoints. This happens in time-invariant electric fields. When talking about electrostatic potential energy, time-invariant electric fields are always assumed so, in this case, the electric field is conservative and Coulomb's law can be used.
Using Coulomb's law, it is known that the electrostatic force F and the electric field E created by a discrete point charge Q are radially directed from Q. By the definition of the position vector r and the displacement vector s, it follows that r and s are also radially directed from Q. So, E and ds must be parallel:
Using Coulomb's law, the electric field is given by
and the integral can be easily evaluated: (en)
- The electrostatic potential energy UE stored in a system of two charges is equal to the electrostatic potential energy of a charge in the electrostatic potential generated by the other. That is to say, if charge q1 generates an electrostatic potential V1, which is a function of position r, then
Doing the same calculation with respect to the other charge, we obtain
The electrostatic potential energy is mutually shared by and , so the total stored energy is
This can be generalized to say that the electrostatic potential energy UE stored in a system of n charges q1, q2, …, qn at positions r1, r2, …, rn respectively, is: (en)
- ∇ × E (en)
- Using the formula given in , the electrostatic potential energy of the system of the three charges will then be:
Where is the electric potential in r1 created by charges Q2 and Q3, is the electric potential in r2 created by charges Q1 and Q3, and is the electric potential in r3 created by charges Q1 and Q2. The potentials are:
Where r'ij is the distance between charge Q'i and Qj.
If we add everything:
Finally, we get that the electrostatic potential energy stored in the system of three charges: (en)
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