# Energy Density in Electrostatic Fields

To determine the Energy present in an assembly of charges (or) group of charges one must first determine the amount of work necessary to assemble them.

It is seen that , when a unit positive charge is moved from infinity to a point in a field, the work is done by the external source and energy is expended.

If the external source is removed then the unit positive charge will be subjected to a force exerted by the field and will be moved in the direction of force.

Thus to hold the charge at a point in an electrostatic field, an external source has to do work , this energy gets stored in the form of Potential Energy when the test charge is hold at a point in a field.

when external source is removed , the Potential Energy gets converted to a Kinetic Energy.

In order to derive the expression for energy stored in electrostatics (i.e, the expression of such a Potential Energy)

Consider an empty space where there is no electric field at all, the Charge $Q_{1}$ is moved from infinity to a point in the space ,let us say the point as $P_{1}$, this requires no work to be done to place a charge $Q_{1}$ from infinity to a point $P_{1}$ in empty space.

i.e, work done = 0 for placing a charge $Q_{1}$ from infinity to a point $P_{1}$ in empty space.

now another charge $Q_{2}$ has to be placed from infinity to another point $P_{2}$ . Now there has to do some work to place $Q_{2}$ at $P_{2}$ because there is an electric field , which is produced by the charge $Q_{1}$ and $Q_{2}$ is required to move against the field of $Q_{1}$.

Hence the work required to be done is  $Potential=\frac{work&space;done}{unit&space;charge}$

i.e, $V&space;=&space;\frac{W}{Q}$ $\Rightarrow&space;W&space;=&space;V&space;X&space;Q$ .

$\therefore$ Work done to position $Q_{2}$ at $P_{2}$ = $V_{21}&space;X&space;Q_{2}$.

Now the charge $Q_{3}$ to be moved from infinity to $P_{3}$ , there are electric fields due to $Q_{1}$ and $Q_{2}$, Hence total work done is due to potential at $P_{3}$ due to charge at $P_{1}$ and Potential at $P_{3}$ due to charge at $P_{2}$.

$\therefore$ Work done to position $Q_{3}$ at $P_{3}$ = $V_{31}Q_{3}+V_{32}Q_{3}$.

Similarly , to place a  charge $Q_{n}$ at $P_{n}$ in a field created by (n-1) charges is ,work done to position $Q_{n}$ at $P_{n}$$=V_{n1}Q_{n}+V_{n2}Q_{n}+V_{n3}Q_{n}+.......$

$\therefore$Total Work done $W_{E}&space;=Q_{2}V_{21}+Q_{3}V_{31}+Q_{3}V_{32}+Q_{4}V_{41}+Q_{4}V_{42}+Q_{4}V_{43}+....&space;EQN(I)$

The total work done is nothing but the Potential energy in the system of charges hence denoted as $W_{E}$,

if charges are placed in reverse order (i.e, first $Q_{4}$ and then $Q_{3}$ and then  $Q_{2}$  and finally $Q_{1}$ is placed)

work done to place $Q_{3}&space;\Rightarrow&space;V_{34}Q_{3}$

work done to place $Q_{2}&space;\Rightarrow&space;V_{24}Q_{2}+V_{23}Q_{2}$

work done to place $Q_{1}&space;\Rightarrow&space;V_{14}Q_{1}+V_{13}Q_{1}++V_{12}Q_{1}$

Total work done $W_{E}&space;=Q_{3}V_{34}+Q_{2}V_{24}+Q_{2}V_{23}+Q_{1}V_{14}+Q_{1}V_{13}+Q_{1}V_{12}+....&space;EQN(II)$

EQN (I)+EQN(II) gives

$2W_{E}&space;=Q_{1}(V_{12}+V_{13}+V_{14}+....+V_{1n})&space;+Q_{2}(V_{21}+V_{23}+V_{24}+....+V_{2n})+Q_{3}(V_{31}+V_{32}+V_{34}+....+V_{3n})+.....$

let $V_{1}=(V_{12}+V_{13}+V_{14}+....+V_{1n})$, $V_{2}=(V_{21}+V_{23}+V_{24}+....+V_{2n})$ and $V_{n}=(V_{n1}+V_{n2}+V_{n3}+....+V_{nn-1})$ are the resultant Potentials due to all the charges except that charge.

i.e, $V_{1}$ is the resultant potential due to all the charges except $Q_{1}$.

$2W_{E}&space;=&space;Q_{1}V_{1}+Q_{2}V_{2}+Q_{3}V_{3}+......+Q_{n}V_{n}$

$W_{E}&space;=\frac{1}{2}&space;\sum_{m=1}^{n}Q_{m}V_{m}$ Joules.

The above expression represents the Potential Energy stored in the system of n point charges.

simillarly,

$W_{E}&space;=&space;\frac{1}{2}\int_{l}\rho&space;_{l}dl.&space;V$  Joules

$W_{E}&space;=&space;\frac{1}{2}\int_{s}\rho&space;_{s}ds.&space;V$ Joules

$W_{E}&space;=&space;\frac{1}{2}\int_{v}\rho&space;_{v}dv.&space;V$ Joules  for different types of charge distributions.

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