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ELECTROSTATICS
CHAPTER ONE: COULOMB’S LAW
Introduction
Charge is the basic essence of electricity. It is the quantity of electricity present in a body.
The quantity of positive charge (due to electrons) in an atom is always equal to the quatity of
positive charge( due to protons), as long as the atom exists in its natural state.
Wherever we separate positive charge from negative charge, we say that we have charged a
body. More specifically charging involves transfer of electrons to or from a body.
A body will be positively charged when it loses electrons, just as a glass rod becomes
positively charged when rubbed with silk.
Conversely, when a body gains electrons it gets positively charged, just as the ebonite rod
gets positively charged on rubbing with wool.
We may charge a body in three ways, namely
i.
By friction
ii.
By induction
iii.
By contact or conduction
Principle of quantization of charge
The charge possessed by a body is always an integral multiple of the electronic charge.
Mathematically,
Q = ne,
…………………………………………………………….(1)
where e is the charge of an electron and n is the number of electrons lost or gained.
However,this principle is of no relevance when we are dealing with macroscopic quantities
of charge since the charge of an electron is very small. ( only 1.6 X10-19C)
Configuration of Charge
Charges may exist in the following configurations:
i.
ii.
iii.
Point Charge: This is a localization of static charge at a single point in space.
Line Charge Distribution: This is an even arrangement of point charges along a
straight or curved line segment. The charge per unit length, denoted by the symbol λ
and measured in Coulomb per metre (C/m), is a characteristic of the distribution
Surface Charge Distribution: This is a uniform distribution of charge over a flat or
curved surface. The descriptive parameter of the distribution is the charge per unit
area,denoted by the symbol σ, and measured in Coulomb per squared metre, (Cm-2)
iv.
Three-dimensional charge distribution or Solid Charge: This is a cluster of
charges in three – dimensional space. The distribution can be described by the charge
per unit volume, denoted by the symbol ρ and measured in Coulomb per cubic
metre, (Cm-3)
Coulomb’s Law
The electrostatic force between two point charges is directly proportional to the product
of their magnitudes and inversely proportional to the square of the distance between them.
Mathematically, if we represent the magnitudes of two point charges by q1 and q2 and the
distance between them by r, then electrostatic force between them is given by:
F=
1
𝑞1 𝑞2
4𝜋𝜀0
𝑟2
……………………………………..(2)
ε0 is called the permittivity of free space or vacuum. It is a constant which determines the
degree to which electrical charges will influence each other in a medium. It is least for free
space and several times larger in other media.
Remarks on Coulomb’s law
i.
The force expressed by Coulomb’s law is bilateral (acts between two bodies) and
therefore obeys Newton’s third law of motion.
ii.
If we wish to write Coulomb’s law in vector form, then we must generalize about the
position of two point charges.
Suppose q1 and q2 are located at points with position vectors r1 and r2 respectively,
The distance between the charges, expressed in vector form, is
r = |⃗⃗⃗
𝑟1 − ⃗⃗⃗
𝑟2 | = |⃗⃗⃗
𝑟2 − 𝑟⃗⃗⃗1 |
1
If we assume that q1 and q2 are like charges then the force on q1 is,
⃗⃗⃗
𝐹1 =
1
𝑞1 𝑞2
4𝜋𝜀0
|𝑟
⃗⃗⃗⃗1 −𝑟
⃗⃗⃗⃗2 |2
, …………………………………..(3)
and the force on q2 is
⃗⃗⃗
𝐹2 =
1
𝑞1 𝑞2
4𝜋𝜀0
|𝑟
⃗⃗⃗⃗2 −𝑟
⃗⃗⃗⃗1 |2
, ……………………………………(4)
Superposition Principle
When a point charge is surrounded by a number of point charges. The electrostatic force
experienced is the vector addition of the forces exerted on it by each of surrounding
charges.
Exercise 1: Charges of +50µC, -50 μC and +50μC are placed at the vertices of an equilateral
triangle ABC . The length of each side of the triangle is 2cm. Calculate the net force on the
charge at B.
The charge at B is attracted by charges at A & C (Unlike charges attract). In the vector
parallelogram for adding the attractive forces is as shown above.
Applying Coulomb’s law
Force on B due to charge at A= FA =
Force on B due to charge at C = FC =
1
𝑞𝐴 𝑞𝐵
4𝜋𝜀0
𝐴𝐵2
1
𝑞𝐴 𝑞𝐶
4𝜋𝜀0
𝐴𝐵2
=
=
Clearly, FA = FC
By paralellogram law, the resultant force is given by
F=√𝐹𝐴 2 + 𝐹𝐶 2 + 2𝐹𝐶 𝐹𝐴 cos 600
= √2𝐹𝐴 2 + 2𝐹𝐴 2 cos 600
= √3𝐹𝐴 2
= F A √3
9𝑋109 𝑋50𝑋50𝑋10−12
{(2𝑋2)𝑋10−4 }2
9𝑋109 𝑋50𝑋50𝑋10−12
{(2𝑋2)𝑋10−4 }2
acting along BA
acting along BC
= 9𝑋109 𝑋
25
4
𝑋10−6 𝑋√3
=97,425 N.
acting along the angle bisector of angle B .
Exercise 2. Charges 10 μC and -20 μC are placed 100cm apart along a straight line as shown
below
At what position (s) must we keep another point charge so that the entire system is in
equilibrium. Determine the magnitude and type of charge.
Dielectric Constant of a medium
The dielectric constant of a medium is the ratio between the electrostatic force acting two
point charges situated at a certain distance apart in vacuum and the electrostatic force
between the same point charges situated at the same distances apart in that medium.
Let q1 and q2 be kept at distance ‘d’ apart, first in vacuum and then in a medium with
permittivity ε0.
Then, the dielectric constant is given by
κ=
κ=
𝐹𝑉𝐴𝐶
𝐹𝑀𝐸𝐷
𝜀
𝜀0
=
1
𝑞1 𝑞2
4𝜋𝜀0 𝑑2
1 𝑞1 𝑞2
4𝜋𝜀 𝑑2
,………………………………………(5)
where ε is the permittivity of the medium.
Additional exercise
1.
Define 1 Coulomb of charge on the basis of coulomb’s law
-1mk
2.
Two point charges having magnitudes q1 and q2 are situated in free space. What is the
nature of the force between charges if (a) q1q2 < 0, (b) q1q2 > 0
3.
The sum of the magnitudes of two point charges is 7 microcoulomb (μC). When
situated 3cm apart, the charges repel each other with a force of 1N. Assuming the
charges are in air, calculate their magnitudes.
Answers
1.
One coulomb is the quantity of charge which, when placed in vaccum at a distance of
1m from an equal and similar charge would repel it with a force of 9 x 109 N.
2.
If q1 q2 > 0, the charges are of the same sign. The force is repulsive because if the
charges are positive, they will repel and if they are negative, then also they will repel.
The negatives will multiply to become positive.
ii.
If q1 q2 < 0, charges will be one positive and one negative and the force is attractive,
since unlike charges attract..
3.
q1 + q2 = 7, taking each charge in µC. using Coulomb’s law,
F=
1
𝑞1 𝑞2
4𝜋𝜀0
𝑟2
r = 3cm =3X 10-2m;
F= 1 N;
1=
q1 = x µC, q2 = (7-x) µC ;
1
4𝜋𝜀0
= 9 X 109 Nm2C-2
9 X 109 x (7 –x) X 10−12
(3𝑋10−2 )2
9 X 10-4 = 9 X 10-3 x(7-x)
x(7-x) = 10
x2-7x +10 =0
(x-5) (x-2) = 0
x = 5cm or x = 2cm.
………………………………………………………………………………………………….
CHAPTER TWO:
ELECTRIC FIELD
Introduction
The term electric field is defined primarily and qualitatively as a region of space where
electrical influence can be experienced.
Mathematically, the electric field is also used as a short form of electric field strength ,
electric field intensity or electric flux density.
The strength of an electric field is defined as the force acting on a unit positive test charge
situated at a point in an electric field.
The unit of field strength is newton per coulomb,(N/C)
Mathematically, electric field strength can be written in the form of a limit, i.e.
lim
𝐹
𝑄→0 𝑄
………………………………………………… (6)
Where Q is a vaninshingly small, positive test charge.
Electric field strength around a positive point charge
Let a positive point charge q be located at a point O in space. Let P be a point situated
at a distance ‘r’ from O. If a positive test charge of magnitude Q is located at P it will
experience a force F given by
F=
1
𝑄𝑞
4𝜋𝜀0
𝑟2
The electric field strength will therefore be of magnitude E, given by
E=
E=
lim
𝐹
𝑄→0 𝑄
1
𝑞
4𝜋𝜀0
𝑟2
= lim
1
4𝜋𝜀0
𝑄𝑞
𝑟2
𝑄
𝑄→0
…………………………………………..(7)
Remarks about a test charge
i.
A test charge is purely hypothetical abstraction and not a real charge.
ii.
It is a charge which is capable of experiencing force but incapable of exeting force.
iii.
It is vanishingly small so that it does not disturb the electric field being tested in any
way.
iv.
It may have any magnitude but for the purpose of defining electric field strength, it is
convenient to assignto it a magnitude of unity.
Electric field due to a dipole
An electric dipole is a system of two equal and opposite point charges separated by a finite
distance.
The size of a dipole is measuring by a vector quantity called electric dipole moment (𝑝) .
𝑝
=
⃗⃗⃗⃗ q, …………………………(8)
2𝑎
Where q is the magnitude of either charge and 2a is the distance between them.
Conventionally, the dipole moment vector is assumed to act from the negative charge to
the positive charge.
Field along the axial line of a dipole
Consider an electric dipole consisting of equal and opposite point charges of magnitude q
separated by distance 2a as shown below:
Let P be a general point along the line joining the positions of the charges. At P the electric
field E is a superposition of two electric field vectors EA and EB such that
E = E B – EA ,
where EB= field at p due to +q located at point B and
EA = field at P due to –q located at point A
Now, EB =
1
𝑞
4𝜋𝜀0
𝑃𝐵2
Similarly , EA =
=
1
𝑞
4𝜋𝜀0
𝐴𝑃2
1
𝑞
4𝜋𝜀0
(𝑟−𝑎)2
=
1
𝑞
4𝜋𝜀0
(𝑟+𝑎)2
acting to the right.
Since PB< PA, EB > EB
Eaxial = EB - EA
𝑞
=
𝑞
=
{
4𝜋𝜀0
𝑞
=
=
{
4𝜋𝜀0
4𝜋𝜀0
𝑞
4𝜋𝜀0
{
{
1
(𝑟−𝑎)2
−
1
}
(𝑟+𝑎)2
(𝑟+𝑎)2 −(𝑟−𝑎)2
(𝑟−𝑎)2 (𝑟+𝑎)2
}
(𝑟+𝑎+𝑟−𝑎)(𝑟+𝑎−𝑟+𝑎)
(𝑟 2 −𝑎 2 )2
2𝑟𝑋2𝑎
(𝑟 2 −𝑎2 )2
}
}
Taking 2aq = p, where p is the magnitude of the dipole moment vector
Eaxial =
1
2𝑝𝑟
4𝜋𝜀0
(𝑟 2 −𝑎2 )2
…………………………………………..(9)
For a short dipole , r≫ a and (r2 – a2) ≈ r2 . equation (9) reduces to
Eaxial =
1
2𝑝
4𝜋𝜀0
𝑟3
………………………………………………….(10)
Important Remarks
i.
Equations (9) and (10) are not applicable at the mid – point O of the dipole. This is
because these equations gives zero field at O but a close inspection shows that the
field is non zero at every point along the line AB.
ii.
The electric field along the axis of a dipole acts parallel to the dipole moment vectors
at points outside the line segment AB. Within the segment AB the axial fields acts
opposite to the dipole moment vector.
Field along the equatorial axis
Let point P located at a distance r from the midpoint of an electric dipole as shown above.
The electric field contributed at P by the positive charge at B is
EB =
1
𝑞
4𝜋𝜀0
𝑃𝐵2
=
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
acting along BP.
Similarly, the field contributed at P due to the negative charge at A is
EA=
1
𝑞
4𝜋𝜀0
𝑃𝐴2
=
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
acting along P
Since ∟PAB = ∟PBA = ,
Then EA and EB are inclined at an angle 2θ. Since PA = PB, EA = EB and the resultant E acts
at angle  with either of EA and EB.
 Net field along the equatorial line acts parallel to line joining the two charges, in opposite
direction to the dipole moment vector
Using the parallelogram law, the net field is,
E = √𝐸𝐴 2 + 𝐸𝐵 2 + 2𝐸𝐴 𝐸𝐵 cos2 𝜃
= √{
=
=
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
E=
√2(2𝑐𝑜𝑠 2 𝜃)
√4𝑐𝑜𝑠 2 𝜃
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
But cos θ =
E =
} 2 {(2 + 2cos2θ)}
2 cos 𝜃
𝑎
1
(𝑟 2 +𝑎2 )2
1
𝑞
4𝜋𝜀0
𝑟 2 +𝑎2
2
𝑎
1
(𝑟 2 +𝑎2 )2
Since 2aq = p, the dipole moment,
E =
1
𝑝
4𝜋𝜀0
(𝑟 2 +𝑎2 )2
3
……………………………………..(11)
For a short dipole, r2 ≫ a2 and the equation reduces to
E=
1
𝑝
4𝜋𝜀0 𝑟 3
…………………………………………(12)
Electric field at any general point around a dipole
Let P be a general point situated at a distance ‘r’ from the midpoint O of an electric dipole as
shown below
The electric field at P can be calculated by resolving the dipole moment vector along the axis
PO and perpendicular to the axis as shown below
The point P lies on the axial line of the dipole component pCos. The electric field
contribution at P by pCos component will be
E1 =
1
2𝑝𝑐𝑜𝑠𝜃
4𝜋𝜀0
𝑟3
Similarly, P lies on the equatorial axis of the dipole component pSin, the electric field
contributed by will be,
E2 =
1
𝑝𝑠𝑖𝑛𝜃
4𝜋𝜀0
𝑟3
Since these vectors are perpendicular, the net field at P is
E = √𝐸1 2 + 𝐸2 2
1
E = √(
4𝜋𝜀
E = √(
E=
0
1
4𝜋𝜀0
1
𝑝
4𝜋𝜀0
𝑟3
2𝑝𝑐𝑜𝑠𝜃 2
)
𝑟3
+ (
𝑝 2
) (4𝑐𝑜𝑠 2 𝜃
𝑟3
1
4𝜋𝜀0
𝑝𝑠𝑖𝑛𝜃 2
)
𝑟3
+ 𝑠𝑖𝑛2 𝜃)
√3𝑐𝑜𝑠 2 𝜃 + 1 ………………………………………(13)
Electric Field Lines
An electric field line is an imaginary line which shows the direction of electric field vector at
any given point in an electric field. They also indicate the path in which a positive charge
would travel in an electric field if it is left free.
Properties of Electric Field Lines
i.
They do not intersect
Reason:
a. If field lines are allowed to intersect two tangents would be possible at the point
of intersection. This implies that the same electric field vector acts in two
different directions, which is meaningless.
b. Alternatively, we can also argue that intersecting field lines provides two
different paths for the same free positive charge. Since a charge particle cannot
travel simultaneously in two directions, electric field lines cannot intersect.
ii.
Electric field lines arise from a surface or strike a surface perpendicularly.
Reason
If the field lines strike a surface at angles other than 900 there will be non zero electric
field along the charged surface. This implies that charges on the surface will have a
tendency to move and create surface currents. The ideas of such currents totally
negate the idea of electrostatics where charges must remain at rest.
For this to occur the field lines must be perpendicular to the surface
iii.
Electric fields originate from positive and terminate at negative charge
Reason
Since a free positive charge will naturally be repelled by a positive charge and
attracted by negative charge. Field lines must show the same course of travel.
iv.
Field lines contract longitudinally exert pressure laterally
v.
Electric field lines are close where the electric field is strong and are far apart when
the electric field is weak.
vi.
Electric field lines do not pass through a material surface.
Sketch the electric field lines of the following
1.
Isolated positive charges
2.
3.
4.
5.
6.
Isolated negative charges
Dipole
Pair of equal positive charges
Pair of equal negative charges
Flat charge surface
Electric field due to circular loop of charge
Consider a circular loop of radius ‘a’ carrying uniformly distributed charge of density of ‘ λ’
Cm-1 .Let dl be a small portion of the circumference of the ring located at a distance r from
P, where P is any general point along the central axis of the ring located at distance x from
O.
The electric field contributed at P by the elemental charge λdl is
dE =
1
𝜆𝑑𝑙
acting away from the element.
4𝜋𝜀0 𝑟 2
We can resolve dE in 2 directions, namely dEcosθ along OP and dEsinθ perpendicular to
OP.
For every element dl considered there will be a diametrically opposite element whose field
contribution at P will also produce components dE Cosθ along OP and dE sinθ normal to
OP.
It is clear that the axial components dEcosθ will add up while the perpendicular components
will cancel out in pairs. Hence, the total field at E is,
E = ∮ 𝑑𝐸𝑐𝑜𝑠𝜃
Since cosθ =
E=∮
E=
E=
𝑥
𝑟
1
𝜆𝑑𝑙 𝑥
4𝜋𝜀0
𝑟2 𝑟
1
𝜆
4𝜋𝜀0
𝑟3
1
𝜆
4𝜋𝜀0
𝑟3
∮ 𝑑𝑙
x 2πa
But 2πaλ = the total charge on the entire loop . Taking this as q,
Also , by Pythagora’s theorem,
1
r = (𝑎2 + 𝑥 2 )2
 E=
1
𝑞𝑥
4𝜋𝜀0
(𝑎2 +𝑥 2 )2
3
…………………………………………(14)
At the centre of the loop
E=O
When the loop is small compared to distance x,
a 2 + x2 ≈ x 2
Equation (14) reduces to
E=
1
𝑞
4𝜋𝜀0
𝑥2
…………………………………………………(15)
This shows that a small loop behaves as a point charge.
Solid angle
The solid angle subtended at a distance r by an area element dS is given by
dΩ =
𝑑𝑆𝑐𝑜𝑠𝜃
𝑟2
Where θ is the angle between the normal to surface element and radially outward vector
from O. Using vectors,
dΩ =
⃗⃗⃗⃗⃗
𝑑𝑆.𝑟
𝑟2
If point O is at the centre of a sphere of radius r and area ds is a part of the surface of the
sphere. Then we have,
dΩ =
𝑑𝑆𝑐𝑜𝑠𝜃
𝑟2
=
𝑑𝑆
𝑟2
This because the normal to the surface of a sphere acts along the radius, making angle θ to be
zero.Hence, the total solid angle subtended at the centre of the sphere will be,
Ω= ∮
𝑑𝑆𝑐𝑜𝑠𝜃
𝑟2
=
1
𝑟2
∮ 𝑑𝑆 =
1
𝑟2
𝑋 4𝜋𝑟 2 = 4π
Hence, 4𝝅 is the measure of a complete solid angle
In general,
∮
𝑑𝑆𝑐𝑜𝑠𝜃
= 4π
𝑟2
Proof of Gauss’s theorem
Consider a surface of arbitrary shape enclosing charge q at some point inside it.
The electric flux through a small area dS is
d = EdS cosθ
Where E = field at distance r from the point charge, i.e.
E=
1
𝑞
4𝜋𝜀0
𝑟2
 d =
1
𝑞
4𝜋𝜀0
𝑟2
dS cosθ
The total flux through the entire surface is found by evaluating the integral
=∮
=
1
𝑞
4𝜋𝜀0
𝑟2
𝑞
∮
4𝜋𝜀0
But ∮
=
𝑑𝑆 cos 𝜃
𝑟2
𝑑𝑆 cos 𝜃
𝑞
4𝜋𝜀0
dS cos θ
𝑟2
= 4π. Hence
X 4π =
𝑞
𝜀0
. Therefore ,
𝑞
𝑑𝑆 =
…………………………………………………..(16)
∮ 𝐸⃗ . ⃗⃗⃗⃗
𝜀
0
Guassian Surface
This is an imaginary surface enclosing charge to which Gauss’s theorem can be applied.
However, we must carefully define a gaussian surface so that the mathematical effort
required for its application is minimal.
Two points are necessary to achieve this namely,
i.
ii.
The surface must be selected such that the magnitude of electric field vector (𝐸⃗ ) does
not vary over the entire Gaussian surface.
It must also be ensured that the direction of the electric field vector is everywhere
perpendicular to the Gaussian surface.
Applications of Gauss’s theorem
1.
Field due to a point charge
Let a point charge q be located at a point O in free space. To calculate the electric
field at distance r, we imagine a spherical Guassian surface centred at O and having
radius r.
By Gaussian’s Theorem,
𝑞
𝑑𝑆 =
∮ 𝐸⃗ . ⃗⃗⃗⃗
𝜀
0
Since the electric field vector from the point charge is radially outwards, it acts parallel to the
area vector dS.
𝑑𝑆 = ∮ 𝐸𝑑𝑆 𝑐𝑜𝑠𝜃 = ∮ 𝐸 𝑑𝑆 cos 00 = ∮ 𝐸 𝑑𝑆
∮ 𝐸⃗ . ⃗⃗⃗⃗
Since every portion of the Gaussian surface is at the same distance r from the point charge,
the magnitude of electric field will be constant. Therefore
∮ 𝐸 𝑑𝑆 = E ∮ 𝑑𝑆
= E X 4π𝑟 2
Hence,
E X 4π𝑟 2 =
E=
1
𝑞
4𝜋𝜀0
𝑟2
𝑞
𝜀0
………………………………………………………………………………………………
Homework
1.
Use Gaussian’s theorem to prove Coulomb’s law.
2.
A point charge (q) is enclosed by an imaginary which surface of side l. Determine the
electric flux through any one face of the cube.
…………………………………………………………………………………………………
Electric field due to line charge distribution
Consider a uniform distribution of charges point of along a straight line as shown below.
We assume the length of the line is infinite.
To calculate the electric field at a distance r from the line charge we define a cylindrical
Gaussian surface with the line charge at its axis and also having radius (r).
For a small portion of area dS of the surface, the area vector ⃗⃗⃗⃗⃗
𝒅𝑺 acts radially outwards and
parallel to electric field vector.
Furthermore, every part of the Gaussian surface is located at the same distance r from the
line charge. Hence, the magnitude of electric field vector is constant.
Applying Gauss’s theorem,
𝑞
⃗⃗⃗⃗ =
∮ 𝐸⃗ . 𝑑𝑆
𝜀
0
Where q = charge enclosed within the Gaussian surface
But the charge per unit length is λ. Hence ,
q = λl
Where l is the length of cylinder.
Evaluating the integral using the conditions given above,
⃗⃗⃗⃗ = ∮ 𝐸 𝑑𝑆 = E∮ 𝑑𝑆 = E X curved surface area of the Gaussian cylinder.
∮ 𝐸⃗ . 𝑑𝑆
= E X 2πrl.
 E X 2πrl = λl
E=
𝜆
2𝜋𝜀0 𝑟
…………………………………………………..( 17)
The result is illustrated graphically below:
problem
An electron is revolving around a linear charge distribution under the influence of
electrostatic attraction. Taking the radius of the orbit as r,derive an expression for the kinetic
energy of the electron.
…………………………………………………………………………………….
3.
Field due to a charged spherical shell
Consider a sphere of radius of radius R carrying uniform charge all over its outer surface.
Let the charge per unit per area be σ
To calculate the electric field at any distance r from the centre of the sphere we have two
cases.
i.
Where r < R
ii.
Where r > R
Case (i) Points lying inside the sphere
Choosing a spherical Gaussian surface concentric with the charge sphere and having radius
r,we find that the Gaussian surface does not enclose any charge, since there is no charge
inside the sphere
𝑑𝑆 = 0
∮ 𝐸⃗ . ⃗⃗⃗⃗
E = 0.
Case (ii): Points on or outside the sphere
For this case we choose a spherical Gaussian surface completely enclosing the charged
sphere.
Applying Gauss’s theorem to the outer sphere,
𝑞
⃗⃗⃗⃗ = ,
∮ 𝐸⃗ . 𝑑𝑆
𝜀
0
Where q is the total charge enclosed by the inner sphere.
For any small portion dS of the Gaussian surface, the area vector ⃗⃗⃗⃗
𝑑𝑆 acts along the common
radius of the two spheres and therefore parallel to the electric field vector 𝐸⃗ radiating
outwards from the surface of the charged surface. Hence ,
𝐸⃗ . ⃗⃗⃗⃗
𝑑𝑆 = E dS cos θ = E dS.
Since every part of the Gaussian surface is at the same distance (r – R) from the surface of
the charged sphere, the electric field vector will be of constant magnitude all over the
surface. Therefore ,
𝑞
𝑑𝑆 = ∮ 𝐸 𝑑𝑆 = E X 4 π 𝑟 2 = 2
∮ 𝐸⃗ . ⃗⃗⃗⃗
𝑟
E=
1
𝑞
4𝜋𝜀0 𝑟 2
…………………………………………………………(18)
This result shows that a charged sphere behaves as if all its charge were concentrated at
its centre.
However, the equation applies only for points on or outside the charged sphere and not in the
inside.
Field due to a solid charged sphere
Consider a sphere of radius R having a three-dimensional (solid) charge distribution of
density ρ Cm-3.
The electric field strength due to the distribution can be calculated at points inside and
outside the sphere.
Field inside a solid sphere
Consider a spherical Gaussian surface concentric with a charged sphere having a radius r less
than R.
For any small part dS of this surface, the electric field vector 𝐸⃗ acts parallel to the area
vector ⃗⃗⃗⃗
𝑑𝑆 since they are both radially outwards.
By the symmetry of the sphere, the electric field vector will have a constant magnitude all
over the surface. Applying Gaussian theorem,
⃗⃗⃗⃗ =
∮ 𝐸⃗ . 𝑑𝑆
𝑞
𝜀0
i.e. ∮ 𝐸 𝑑𝑆 =
E ∮ 𝑑𝑆 =
𝑞
𝜀0
,
𝑞
𝜀0
𝑞
E X 4π𝑟 2 =
𝜀0
Where q = charge enclosed within the Gaussian surface
i.e
q =
4
3
E = 4 π 𝑟2 =
E=
𝜌
3𝜀0
𝜋𝑟 3 𝜌
4
3𝜀0
𝜋 𝑟3𝜌
r …………………………………………………………..(19)
Case (iii) field at point outside the sphere
The Gaussian surface is spherical and concentric but, completely enclosing it, i.er >>R
Using Gauss’s theorem,
𝑑𝑆 =
∮ 𝐸⃗ . ⃗⃗⃗⃗
𝑞
𝜀0
,
Where q is the charge enclosed by the solid sphere. Since 𝐸⃗ is everywhere parallel to ⃗⃗⃗⃗
𝑑𝑆 and
of constant magnitude all over the Gaussian surface,
⃗⃗⃗⃗ =
∮ 𝐸⃗ . 𝑑𝑆
∮ 𝐸 𝑑𝑆 = E X 4π r2
 E X 4π r2 = =
E=
1
𝑞
𝑞
𝜀0
……………………………………………………………(20)
4𝜋𝜀0 𝑟 2
This again shows that a solid charged sphere approximates to a point charge
concentrated at its own centre at points on or outside the sphere.
A more detailed expression can be obtained by writing,
q=
4
3
E =
E =
π
1
𝑅3
𝑟2
ρ
4
4𝜋𝜀0 3
𝜌
𝑅3
3𝜀0 𝑟 2
π
𝑅3
𝑟2
ρ
……………………………………………………….(21)
We can illustrate the field variation graphically as shown below,
Field due to a plane charge distribution
Consider a uniform distribution of charge on an infinitely large flat surface.
The electric field at any given distance from the plane can be calculated by taking a Gaussian
surface which encloses a part of the infinite charge.
For convenience, we select a uniform cylinder whose axis is perpendicular to the plane
charge such that it is bisected by the plane charge.
It is clear that the electric field lines perpendicular to field lines are perpendicular to the
circular ends of the Gaussian cylinder. Thus, everywhere on the Gaussian surface, the
electric field vector acts parallel to the area vector.
Furthermore, since each circular end is at the same distance from the plane charge, the
magnitude of the electric field vector is the same.
𝑑𝑆 =
∮ 𝐸⃗ . ⃗⃗⃗⃗
∮ 𝐸 𝑑𝑆 = E X 2A
Charge enclosed within the surface = σ x A
 E X 2A =
E=
𝜌
2𝜀0
𝜎𝐴
𝜀0
……………………………………………………………….(22)
CHAPTER THREE:
ELECTRIC POTENTIAL
Introduction
The electric potential at a point is defined as the work done to transfer one coulomb of
positive charge from infinity to that point.It is a scalar quantity measured in volts.
Definition :One volt is the potential at a point if the work done to transfer one coulomb of
positive charge from infinity to that point is one joule.
i.e 1 volt = 1 joule per coulomb, or
1 V = 1 JC-1
Conservative nature of electrostatic force
Let a test charge Q be transferred from point A to point B in electric of charge q. The path
from A to B is arbitrary.
Point P is a general point along the path from A to B such that OP = r where O is the
location of charge q.Clearly, the test charge placed at P will experience a force,
F = QE,
where E is the electric field intensity at P.
Let dS be a small displacement of the test charge along the path. The work done by the force
is,
dW = FdS cosθ,
where θ is angle between dS and F
But dS cosθ is the component of dS along the direction of F and r. We can therefore write,
dS cosθ = dr
dW = Fdr
1
But, E =
dW =
= Q E dr
𝑞
4𝜋𝜀0 𝑟 2
1
𝑄𝑞
4𝜋𝜀0
𝑟2
dr
The work done in moving the test charge from A to B is
𝑟
WAB = ∫𝑟 𝐵
1
𝑄𝑞
𝐴 4𝜋𝜀0
WAB = -
𝑄𝑞
4𝜋𝜀0
[
1
𝑟𝐵
𝑟2
−
1
𝑟𝐴
] ………………………………………………..(23)
Clearly, equation (23) shows that the work done in moving a test charge from A to B does
not depend on the shape of the path followed but only on the positions A and B . In other
words,
Electrostatic force is conservative.
Furthermore, if we consider the work done against the electrostatic force, by the agent
transferring the test charge from A to B, it will be opposite to the work done by the electric
field. That is,
WAB =
𝑄𝑞
4𝜋𝜀0
1
[
𝑟𝐵
−
1
𝑟𝐴
] ……………………………………………(24)
If the point A is at infinity and the test charge is 1 unit, that is ,one coulomb, the expression
for WAB transforms, by definition to the electric potential at B.
VB =
1
𝑞
4𝜋𝜀0 𝑟𝐵
……………………………………………………(25)
Generally ,the electric potential at a distance r from a point charge q is given by
V=
1
𝑞
…………………………………………………………..(26)
4𝜋𝜀0 𝑟
The above results show that the potential around a positive point charge is positive and
potential around a negative point charge is negative.
If we look fundamentally at definition of electric field strength and potential we conclude
that
V = -∫ 𝐸 𝑑𝑟
Or
E=-
𝑑𝑉
𝑑𝑟
( integral relationship between electric field and potential)…..(27)
( differential relationship between electric field and potential) ……(28)
Proof
The work done against the electrostatic force F to accomplish a displacement dr is given by
dW = -Fdr,
where the displacement dr is assumed to occur in the direction of the force.
The total work done over a given displacement will then be:
W = − ∫ 𝐹 𝑑𝑟
Dividing oth sides by Q,
𝑊
𝑄
= -∫
𝐹
𝑄
𝑑𝑟
But the quantity
𝑾
𝑸
is, by definition, the electric potential, V while
𝑭
𝑸
is the electric field
strength, E . Hence,
V = -∫ 𝐸 𝑑𝑟
Equipotentials or equipotential surfaces
A equipotential is a surface (real or imaginary) at every point of which electric potential is
the same. It has following properties:
i.
ii.
They are everywhere perpendicular to electric field lines
The value of electric potential on equipotential surfaces decreases in the direction
of electric field lines.
iii.
The work done in transferring a charge between two points on same equipotential
surface is zero.
………………………………………………………………………………………………
Homework
Derive the expression for the electric potential due to
i.
A line charge distribution
ii.
An electric dipole
iii.
A plane charge distribution
…………………………………………………………………………………………………
Potential energy
The electric potential energy of a system of point charges is the sum total of the work done
against the electric field in positioning them in that field.
Consider a system of two point charges q1 and q2 placed at distance r apart. We know that
the potential energy of the system is
W=
1
𝑞1 𝑞2
4𝜋𝜀0
𝑟
………………………………………………………………………(29)
Extending this to a system of three point charges, arranged as shown below, the potential
energy of the system can be obtained by taking the charges two at a time. Hence ,
Total potential energy , W = W12 + W23 + W31
W==
1
4𝜋𝜀0
[
𝑞1 𝑞2
+
𝑟12
𝑞2 𝑞3
𝑟23
+
𝑞3 𝑞1
𝑟31
] ……………………………………………….(30)
Generalising the above result for n point charges, we have
W=
1
2
[=
1
4𝜋𝜀0
𝑖,𝑗=𝑛 𝑞𝑖 𝑞𝑗
}
𝑟𝑖𝑗
……………………………………………………(31)
] ∑𝑖,𝑗=1 {
Electric potential of an electric dipole.
Consider a dipole AB having charges -q at point A and charge +q at point B. Let P be
a point at a distance r from the midpoint of the dipole such that <POB = θ
The electric potential at P due to –q is
V1 = −
1
𝑞
4𝜋𝜀0 𝑃𝐴
Similarly, the electric potential at P due to +q is
V2 = =
1
𝑞
4𝜋𝜀0 𝑃𝐵
The net potential at P will therefore be:
VTOTAL =
𝑞
4𝜋𝜀0
[
1
𝑃𝐵
In ∆ AMO, cos θ =
−
1
𝑃𝐴
𝑂𝑀
𝑎
]
, hence a cosθ = OM.
Also , in ∆ BNO, cos θ =
𝑂𝑁
𝑎
, hence a cosθ = ON
Since we are dealing with a short dipole,
PA ≈PO + OM = r + a cosθ
PB ≈ PO – ON = r – a cosθ
𝑞
V=
V=
V=
4𝜋𝜀0
𝑞
4𝜋𝜀0
𝑞
4𝜋𝜀0
[
[
1
𝑟−𝑎𝑐𝑜𝑠𝜃
1
𝑟+𝑎𝑐𝑜𝑠𝜃
𝑟+𝑎𝑐𝑜𝑠𝜃−𝑟+𝑎𝑐𝑜𝑠
𝑟 2 −𝑎 2 𝑐𝑜𝑠 2 𝜃
2𝑎𝑐𝑜𝑠𝜃
[
−
𝑟 2 −𝑎2 𝑐𝑜𝑠 2 𝜃
]
]
]
Since 2aq = p, the magnitude of the dipole moment vector,
V=
1
4𝜋𝜀0
[
𝑝𝑐𝑜𝑠𝜃
𝑟 2 −𝑎2 𝑐𝑜𝑠 2 𝜃
]
But the dipole is small, so that {𝑟 2 − 𝑎2 𝑐𝑜𝑠 2 𝜃} ≈ r2. Therefore,
V=
1
𝑝𝑐𝑜𝑠𝜃
4𝜋𝜀0
𝑟2
………………………………………….(32)