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 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}+.......

\thereforeTotal 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.

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

2W_{E} = Q_{1}V_{1}+Q_{2}V_{2}+Q_{3}V_{3}+......+Q_{n}V_{n}

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

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


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

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

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

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.