# Energy Associated With A Charge Distribution

Do you know that the energy is associated with a charge distribution . This  associated energy is also known as Electrostatic Energy. Let us calculate this energy associated with charge distribution:

When a test charge is moved from a point of lower potential to a point of higher potential, the potential energy of the system increases. But, there is a decrease in potential energy when a test charge is moved from higher to lower potential.

The potential energy imparted to the system by external work on a charge when moved from lower to higher potential is

W=QV

Where W is the work done, Q is the test charge. To calculate the energy of a system say consisting of three points charges Q,Q2,Q3 ,we first bring the charge Q1 from infinity to position 1,no work is done here as the field does not exist in the system.

Thus                                              W1=0

Similarly for q2, work done ,W2=Q2V21.where v2.1 is the potential at the location of Q2due to Q1.Also for Q3we have W3=Q3V3.1+Q3.V3.2 where V3.1 and V3.2 are potentials at position 3 due to Q and Q2respectively.

Total work done                                          W=W1+W2+W3 (1)

=0+Q2V2V2.1+Q3V3.1+Q3V3.2

If charges are brought in the reverse order then,

W=W3+W2+W1=0+Q2V2.3+Q1V1.3+Q1V1.2 (2)

Adding equations (2) and (3),we get

2W=Q1V1.3+Q1v1.2+Q2V21+Q2V21+Q3V3.1+Q3V3.2 (3)

V1.3+V1.2 is the sum of potential at position Q1due to charges Q3 amd Q2 respectively i.e total potential at Q1 is V1 .Similarly total potential at Q2 ans Q3 is V2 and V3 respectively. Thus equation (3) can be written as

2W=Q1V1+Q2v2+Q3V3

Or                                                          W=1/2 (Q1V1+Q2V2+Q3V3)

For large number of charges ,we have

W=1/2 ∑ni=1QiVi

But for a continuous charge distribution, work done in a volume V having charge density ρv enclosed within a sphere of radius R is

W=1/2∫ρv VdV                                                  (Q=ρvdV)

W=1/2 ∫(del.D)VdV                          [ from Gauss’s law del.D=ρv]         (4)

As per vector identity ,we know that

Del(del D)=(del D)V+D.(delV)

Or                                                               (Del.D)V=(del.VD)-(D.del V)

Put this equation in equation (4)

W=1/2 ∫(del.VD)dV-1/2∫(D.delV)dV                                                             (5)

Now for large R→∞,using divergence theorem

=1/2 ∫(Del.VD)dV =1/2 ∫VD.dS

At very large radius r,D=1/R2,V∞1/R so VD∞1/R3 and surface ∞R2

Thus               ½ ∫VD.dS=0                                             ( 1/2∫(1/R3)R2=0)

From equation (s),we have

W=1/2 ∫(D.delV)dV

Since                                                     E=-del V

W=1/2 ∫(D.E)dV

Also                                                       D=ε0E

Thus ,Energy                                     W=1/2 ∫ε0E2dV

In the above expressions,it appears that each tiny volume dV be assigned the energy content WdV where

Energy density                            =E/dV

WE=1/2 ε0E2

Here                                                    ε=ε0εr

Energy Density,

WE=1/2 ε0εrE2