Energy Density in Electrostatic Fields

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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 $Potential=\frac{work done}{unit charge}$

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

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

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 $W_{E} =Q_{2}V_{21}+Q_{3}V_{31}+Q_{3}V_{32}+Q_{4}V_{41}+Q_{4}V_{42}+Q_{4}V_{43}+.... EQN(I)$

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 $Q_{3} \Rightarrow V_{34}Q_{3}$

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

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

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

EQN (I)+EQN(II) gives

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

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.