The presentation highlights the basics of electrostatic energy and the energy density in the electrostatic field.

1. ENERGY IN THE FIELD
Prepared by:
Hasit Trivedi
hasittrivedi10@gmail.com

2. PRESENTATION OUTLINE
 Electrostatic Field
 Electrostatic Energy
 Energy Density
 Example
 References

3. ELECTROSTATIC FIELD
 The study of interactions of electric charges that are at rest
is called electrostatics.
 Electric field strength (𝐸) is the force that a unit positive charge
experiences when placed in electric field.
 Electric field produced by static charges is called electrostatic
field.

4. ELECTROSTATIC ENERGY
 It is the potential energy of group of charges separated by
some distance from each other in a given system. Thus, it is
called electric potential energy.
 The work done to move an electric charge from point a to point
b in presence of electric field is
(J= Joules)𝑊 = −𝑄 ‫׬‬𝑎
𝑏
𝐸. 𝑑𝑙 J
𝑊 = 𝑄. 𝑉𝑎𝑏 J

5.  It is observed that when a unit positive charge is moved
from infinity to a point in an electric field, the work is done
by the external source against the electric field.
 Thus, work done to move an electric charge from infinity to
a point in the electric field is stored in the form of potential
energy.
 And as soon as the external force is removed this potential
energy is converted into the kinetic energy and charge starts
moving in the direction of the field.

6.  Initially we consider there is no charge in free space.
It means there is no electric field and hence no role of energy
in that electrostatic field.
 Now we consider by external source some charges are moved
into the electric field from infinity.
Figure (i) : assembly of charges

7.  The work done to bring charge 𝑄1 from infinity is
𝑊1 = 0 J
 The work done to bring charge 𝑄2 from infinity is
𝑊2 = 𝑄2 𝑉21 J
 The work done to bring charge 𝑄3 from infinity is
𝑊3 = 𝑄3(𝑉31 + 𝑉32) J
 The work done to bring charge 𝑄4 from infinity is
𝑊4 = 𝑄4(𝑉41 + 𝑉42 + 𝑉43) J

8.  Thus, the total energy stored in the system of charges is
equal to the total work done to move these charges.
 Therefore, total work done is
𝑊𝑇 = 𝑊1 + 𝑊2 + 𝑊3 + 𝑊4 J
𝑊𝑇= 0 + 𝑄2 𝑉21 + 𝑄3 𝑉31 + 𝑉32 + 𝑄4(𝑉41 + 𝑉42 + 𝑉43) J
𝑊𝑇= 𝑄2 𝑉21 + 𝑄3 𝑉31 + 𝑄3 𝑉32 + 𝑄4 𝑉41
+𝑄4 𝑉42 + 𝑄4 𝑉43 J …(1)

9.  Similarly, charge 𝑄 𝑛 is moved from infinity to point 𝑟𝑛 in
presence of electric field produced by (n-1) charges the work is
done against the system.
 Thus, the total work done will be equal to
𝑊𝑇 = 𝑊1 + 𝑊2 + 𝑊3 + 𝑊4 + … … … + 𝑊𝑛 J
𝑊𝑇 = 0 + 𝑄2 𝑉21 + 𝑄3 𝑉31 + 𝑉32 + 𝑄4 𝑉41 + 𝑉42 + 𝑉43 +
… … + 𝑄 𝑛 𝑉𝑛1 + 𝑉𝑛2 + 𝑉𝑛3 + … … + 𝑉𝑛.𝑛−1 J

10.  Now, if the charges were placed in reverse order,
Then, from figure (i)
 The work done to bring charge 𝑄4 from infinity is
𝑊4 = 0 J
 The work done to bring charge 𝑄3 from infinity is
𝑊3 = 𝑄3 𝑉34 J
 The work done to bring charge 𝑄2 from infinity,
𝑊2 = 𝑄2(𝑉23 + 𝑉24) J
 The work done to bring charge 𝑄1 from infinity,
𝑊1 = 𝑄1(𝑉12 + 𝑉13 + 𝑉14) J

11.  Therefore, total work done is
𝑊𝑇 = 𝑊4 + 𝑊3 + 𝑊2 + 𝑊1 J
𝑊𝑇 = 0 + 𝑄3 𝑉34 + 𝑄2(𝑉23 + 𝑉24) + 𝑄1(𝑉12 + 𝑉13 + 𝑉14) J
𝑊𝑇 = 𝑄3 𝑉34 + 𝑄2 𝑉23 + 𝑄2 𝑉24 + 𝑄1 𝑉12 + 𝑄1 𝑉13 + 𝑄1 𝑉14 J …(2)
 Now on adding equations (1) & (2),
we get
𝑊𝑇 = 𝑄1 𝑉12 + 𝑉13 + 𝑉14 + 𝑄2 𝑉21 + 𝑉23 + 𝑉24
+𝑄3 𝑉31 + 𝑉32 + 𝑉34 + 𝑄4(𝑉41 + 𝑉42 + 𝑉43) J
𝑊𝑇 =
1
2
෍
𝑘=1
4
𝑄 𝑘 𝑉𝑘

12.  Similarly, if there are N number of charges present in the
defined system,
 Then the total energy of the system is

13.  Now instead of point charges, suppose the region has a
continuous charge distributions.
𝑊𝑇 =
1
2
න
𝑙
𝜌𝑙 𝑉𝑑𝑙 (for linear charge)
𝑊𝑇 =
1
2
න
𝑠
𝜌𝑠 𝑉𝑑𝑠 (for surface charge)
𝑊𝑇 =
1
2
න
𝑣
𝜌 𝑣 𝑉𝑑𝑣 (for volume charge)
 Thus, this can be generalized in the limit of continuous charge
distribution to an integral over a charge density ρ.

14.  Now utilizing the point form of Gauss law;
‫ׯ‬𝑠
𝐷. 𝑑𝑠 = ‫׬‬𝑣
𝜌 𝑣 𝑑𝑣
𝛻. 𝐷 = 𝜌 𝑣 … (4)
above equation(4) is Maxwell’s first equation
now on replacing in eqn(3) from eqn(4)
we get,
 𝑊𝑇 =
1
2
‫׬‬(𝛻. 𝐷) 𝑉𝑑𝑣 … (5)

15.  Now using vector identity for any vector 𝐷 and scalar V,
𝛻. 𝑉𝐷 = 𝐷. 𝛻𝑉 + 𝑉 𝛻. 𝐷
𝛻. 𝐷 𝑉 = 𝛻. 𝑉𝐷 − 𝐷. 𝛻𝑉 … (6)
 Substituting the eqn(5) by eqn(6), we get
𝑊𝑇 =
1
2
‫׬‬ 𝛻. 𝑉𝐷 𝑑𝑣 −
1
2
‫׬‬ 𝐷. 𝛻𝑉 𝑑𝑣 ….(7)
 Now applying the gauss law to the first term of right hand side
of eqn (7),we get
𝑊𝑇 =
1
2
‫ׯ‬𝑠
𝛻. 𝑉𝐷 𝑑𝑠 −
1
2
‫׬‬𝑣
𝐷. 𝛻𝑉 𝑑𝑣 …(8)

16.  Hence, eqn (8) reduces to
𝑊𝑇 = −
1
2
‫׬‬𝑣
𝐷. 𝛻𝑉 𝑑𝑣 …(10)
 We know that the electric field intensity (𝐸) is the gradient of
electric potential (V).
𝐸 = −𝛻𝑉 … (11)
 Hence, using eqn (11),
we get
𝑊𝑇 =
1
2
‫׬‬𝑣
(𝐷. 𝐸) 𝑑𝑣

17.  Also, as we know
𝐷 = 𝜀 𝑜 𝐸
 Hence, electrostatic energy is
𝑊𝑇 =
1
2
න
𝑣
(𝐷. 𝐸) 𝑑𝑣__
𝑊𝑇 =
1
2
න
𝑣
(𝜀 𝑜 𝐸2
) 𝑑𝑣
𝑊𝑇 =
1
2
න
𝑣
𝐷2
𝜀 𝑜
𝑑𝑣____
 On differentiating above eqns, we get
d = d𝑊𝑇 = 𝐷. 𝐸 𝑑𝑣 = 𝜀 𝑜 𝐸2 𝑑𝑣 =
𝐷2
𝜀 𝑜
𝑑𝑣

18. ENERGY DENSITY
 Energy stored in the field of system of charges per unit
volume is called energy density of the electrostatic field.
𝑤 𝑇 =
d𝑊 𝑇
𝑑𝑣
𝑤 𝑇 = 𝐷. 𝐸 = 𝜀 𝑜 𝐸2 =
𝐷2
𝜀 𝑜
 Thus, the electrostatic energy can written as
𝑊𝑇 = ‫׬‬𝑣
𝑤 𝑇 𝑑𝑣 (𝑖𝑛 𝐽/𝑚3)

19. EXAMPLE
 Question:
The point charges -2nC, 8nC and 6nC are located at (0, 0, 0), (0, 0, 1) and
(1, 0, 0) respectively. Find the energy in the system.
 Solution:
𝑊𝑇 =
1
2
෍
𝑘=1
3
𝑄 𝑘 𝑉𝑘
𝑊𝑇 =
1
2
(𝑄1 𝑉1 + 𝑄2 𝑉2 + 𝑄3 𝑉3) J
𝑊𝑇 =
1
2
𝑄1 𝑉12 + 𝑉13 + 𝑄2 𝑉21 + 𝑉23 + 𝑄2 𝑉31 + 𝑉32 J

20. 𝑊𝑇 =
𝑄1
2
𝑄2
4𝜋𝜀 𝑜(1)
+
𝑄3
4𝜋𝜀 𝑜(1)
+
𝑄2
2
𝑄1
4𝜋𝜀 𝑜(1)
+
𝑄3
4𝜋𝜀 𝑜(√2)
+
𝑄3
2
𝑄1
4𝜋𝜀 𝑜(1)
+
𝑄2
4𝜋𝜀 𝑜(√2)
J
𝑊𝑇 =
1
4𝜋𝜀 𝑜
𝑄1 𝑄2 + 𝑄1 𝑄3 +
𝑄2 𝑄3
√2
J
= 9
48
2
− 28 nJ = 53.48 nJ

21. References
 Principle of electromagnetics by Matthew N.O. Sadiku
 https://www.youtube.com/watch?v=s
QBi0DZYOqY&feature=emb_logo (NPTEL lecture)

22. Thank you