Electrostatics Chapter 3 - جامعة نزوى · •Upon learning the material presented in this...
Transcript of Electrostatics Chapter 3 - جامعة نزوى · •Upon learning the material presented in this...
Electrostatics
Chapter 3
Objectives
• Upon learning the material presented in this chapter,
you should be able to :
1- Evaluate the electric field and electric potential due to any
distribution of electric charges.
2- Apply Gauss’s law.
3- Calculate the resistance 𝑅 of any shaped object, given the
electric field at every point in its volume.
4- Calculate the capacitance of two – conductor configurations.
5- Calculate the electric field at the interface between two media.
3-1 Maxwell’s Equations
• The base of modern electromagnetism is a set
of four fundamental relations known as Maxwell's equations :
1- ∇. 𝐷 = 𝜌𝑣
2- 𝛻 × 𝐸 = −𝜕𝐵
𝜕𝑡
3- 𝛻. 𝐵 = 0
4- 𝛻 × 𝐻 = 𝐽 +𝜕𝐷
𝜕𝑡
Gauss’s law
Faraday’s law.
No monopole (individual pole doesn’t exist).
Ampere’s law
1- 𝛻. 𝐷 = 𝜌𝑣
2- 𝛻 × 𝐸 = −𝜕𝐵
𝜕𝑡
3- 𝛻. 𝐵 = 0
4- 𝛻 × 𝐻 = 𝐽 +𝜕𝐷
𝜕𝑡
𝐸: electric field intensity.
𝐷=𝜀𝐸, 𝐷 is the electric flux density.
𝐵: magnetic flux density .
𝐵=𝜇𝐻 , 𝐻 is the magnetic field intensity.
𝜌𝑣: electric charge density per unit volume.
𝐽 : current density.
Where,
Published in 1873 by
James Clerk Maxwell.
What does each equation mean?!
∇. 𝐷 = 𝜌𝑣
Maxwell’s first equation is Gauss's
law.it says that electric field lines
diverge from positive charges and
converge to negative charges.
𝛻 × 𝐸 = −𝜕𝐵
𝜕𝑡
The second equation is Faraday’s
law. Electric field lines curl
around changing magnetic fields.
Changing magnetic fields induce
electric fields.
𝛻. 𝐵 = 0 The third equation says that
magnetic fields never diverge or
converge. They always go in
closed curves.
𝛻 × 𝐻 = 𝐽 +𝜕𝐷
𝜕𝑡
Finally, the last equation says that
magnetic fields lines curl around
electric currents. We have seen that
a magnetic field circles around a
conducting wire.
• In the static case, non of the quantities appearing in
Maxwell’s equations are functions of time (i.e., 𝜕
𝜕𝑡= 0).
• This happens when all charges are permanently fixed in
space, or, if they move, they do so at a steady rate so that
𝜌𝑣 and J are constant in time.
• Under these circumstances, the time derivatives of
B and D in the second and third equations vanish (=zero),
and Maxwell's equations reduce to:
Electrostatics:
Magnetostatics:
** So in static case ,𝐸 and 𝐵 are no longer interconnected.
∇. 𝐷 = 𝜌𝑣
∇× 𝐸 = 0
∇. 𝐵 = 0
𝛻 × 𝐻 = 𝐽
Only the electric
field quantities
𝐸and 𝐷
Only the
magnetic field
quantities 𝐵 and
𝐻
3-2 Charge and Current Distributions
3-2.1 Charge densities:
• In general ,charge distributions vary with position over the
line, surface or volume and require an integration to
determine the total charge 𝑄 in the distribution.
• The volume charge density 𝜌𝑣 is defined as:
𝜌𝑣 = lim∆𝑣→0
∆𝑞
∆𝑣=𝑑𝑞
𝑑𝑣 𝐶 𝑚3
• The total charge contained in a given volume is given by:
𝑄 = 𝜌𝑣𝑑𝑣 (𝐶)𝑣
,where ∆𝑞 is the charge contained in ∆𝑣.
• The surface charge density 𝜌𝑠 is defined as:
𝜌𝑠 = lim∆𝑠→0
∆𝑞
∆𝑠=𝑑𝑞
𝑑𝑠 𝐶 𝑚2
,where ∆𝑞 is the charge present across an elemental surface area ∆𝑠.
𝑄 = 𝜌𝑠𝑑𝑠 (𝐶)𝑠
• The line charge density 𝜌𝑙 is defined as:
𝜌𝑙 = lim∆𝑙→0
∆𝑞
∆𝑙=
𝑑𝑞
𝑑𝑙 𝐶 𝑚
𝑄 = 𝜌𝑙𝑑𝑙 (𝐶)𝑙
1Example
𝜌𝑙 = 2𝑧 on a cylindrical tube.
If 𝑙 = 10cm,what is the total charge on this tube ?
Q= 𝜌𝑙𝑑𝑙 = 2𝑧𝑑𝑧 = 𝑧20.10
0 = 0.01 C
solution
0.1
0
2Example
The charge on a disk is linearly increasing with r from zero
at center to 6 C/𝑚2at r =3cm.find the total charge on this
disk.
Solution
(𝜌𝑠 ∝ 𝑟 ∴ 𝜌𝑠 = const 𝑟 , const is the slope of the line,
𝑠𝑙𝑜𝑝𝑒 =6 − 0
0.03 − 0=
6
0.03 ∴ 𝜌𝑠=
6𝑟
0.03
𝑄 = 6𝑟
0.03
0.03
𝑟=0
2𝜋
𝜙=0 𝑟𝑑𝑟𝑑𝜙
𝑄 = 𝑑𝜙 200
0.03
0
𝑟2𝑑𝑟 = 2𝜋 × 200 ×𝑟3
3
= 11.31 × 10−3 (𝑐)
2𝜋
𝜙=0
0
0.03
𝑄 = 𝜌𝑠𝑑𝑠𝑠
𝑑𝑠 = 𝑟𝑑𝑟𝑑𝜙
3-2.2 Current Density
• Consider a tube of charge with volume charge density
𝜌𝑣 , as shown in fig 3 − 2 (a).
• The charges are moving with a mean velocity 𝑢 and over a period
∆𝑡, they move a distance ∆𝑙 = 𝑢∆𝑡.
** Now consider the case where the charges are moving through
a surface ∆𝑠 whose surface normal 𝐧 is not parallel to u as shown in Fig 3 − 2(b)
• The amount of charge that crosses the tube’s cross-sectional
surface ∆𝑠′ (whose surface normal 𝒏 is parallel to 𝑢) in time ∆𝑡 is therefore
,where
• The amount of charge ∆𝑞 flowing through ∆𝑠 is
→ →
→ → → →
Where ∆𝑠 = 𝒏 ∆𝑠
• The corresponding current is
𝑱 is called the current density
• The total current 𝐼 through an arbitrary surface
𝑆 is then the flux of the 𝐽 vector through 𝑆:
→ →
*** Conduction current, which will be discussed in more detail in
section 3-7 ,obeys Ohm’s law (R=𝑉
𝐼 ) ,whereas convection current does
not.
3-3 Coulomb’s Law
1- 𝐸 for an isolated charge q
is given by:
𝐸 = 𝑹 𝑞
4𝜋𝜀𝑅2
,where
𝑹 : unit vector pointing from 𝑞 to 𝑃.
𝑅: distance between 𝑞 and 𝑃.
𝜀: electrical permittivity of the medium containing
the observation point. 𝑃.
Coulomb’s law, which was first introduced for electrical charges in
air and later generalized to material medium, states that:
2- The force 𝐹 on a test charge 𝑞’ in an electric field 𝐸
is given by :
𝐹 = 𝑞′𝐸 (N)
• For a material with electrical permittivity 𝜀,
𝐷 and 𝐸 are related by :
𝐷 = 𝜀𝐸
with ε = 𝜀𝑟𝜀0
where 𝜀0 = 8.85 × 10−12 ≅10−9
36𝜋 (𝐹 𝑚 )
is the electrical permittivity of free space, and
𝜀𝑟 = 𝜀 𝜀0 is called the relative permittivity or dielectric
constant.
• For most materials, 𝜀 of the material has a constant value independent of both the magnitude and direction of 𝐸 .
• If ε is independent of the magnitude of E, then the
material is said to be linear because
D and E are related linearly, and if it is independent
of the direction of E , the material is said to be isotropic.
3-3.1 Electric Field due to Multiple Points Charges
• Consider two point charges 𝑞1 and 𝑞2,located at
position vectors 𝑅1 𝑎𝑛𝑑 𝑅2 from the origin of the
given coordinate system, as shown in Fig 3-4.
• In the Eq: 𝐸 = 𝑹 𝑞
4𝜋𝜀𝑅2
Replace 𝑅 with 𝑅 − 𝑅1
And 𝑹 with 𝑅 − 𝑅1
𝑅 − 𝑅1
Thus,
𝐸1 =𝑞1
4𝜋𝜀 𝑅 − 𝑅12
𝑅 − 𝑅1
𝑅 − 𝑅1
𝐸1 =𝑞1(𝑅−𝑅1)
4𝜋𝜀 𝑅−𝑅13 (v/m)
Where,
𝑅 is the vector for point 𝑝
𝑅1 is the vector for the
charge location (𝑞1 location)
• Similarly,
𝐸2 =
𝑞2(𝑅−𝑅2)
4𝜋𝜀 𝑅−𝑅23 (v/m)
• The electric field obeys the principle of linear superposition
𝐸 = 𝐸1 + 𝐸2
𝐸 =𝑞1(𝑅 − 𝑅1)
4𝜋𝜀 𝑅 − 𝑅13 +
𝑞2(𝑅 − 𝑅2)
4𝜋𝜀 𝑅 − 𝑅23
,So
And if we have N charges of 𝑞1, 𝑞2, … . 𝑞𝑁 at locations 𝑅1, 𝑅2, … . 𝑅𝑁, 𝐸
is given by :
𝐸 = 1
4𝜋𝜀
𝑞𝑖(𝑅 − 𝑅𝑖)
𝑅 − 𝑅𝑖3
𝑁
𝑖=1
(𝑉 𝑚 )
𝑁: the total number of charges
𝑞𝑖 : amount of the 𝑖 th charge.
𝑅𝑖 : the position vector of the 𝑖 th charge.
,where
Two point charges with q1 = 2 × 10–5 C and q2 = – 4 × 10–5 C are located at (1, 3, –1)
and (–3, 1, –2), respectively, in a Cartesian coordinate system. Find (a) the electric
field 𝐸 at (3, 1, –2) and (b) the force on a charge q3 = 8 × 10–5 C located at that point.
Solution: (a) Since e = e 0 and there are two charges, we have
with
1 1 2 2
3 30
1 2
1
4
q q
e
R R R RE
R R R R
1
2
ˆ ˆ ˆ3
ˆ ˆ ˆ3 2
ˆ ˆ ˆ3 2
R x y z
R x y z
R x y z
Example
Answer: (a) [continued] After substitution, we find
(b) Using the force equation, we have
5 5 103
0 0
ˆ ˆ ˆ ˆ ˆ ˆ4 2 2 8 48 10 10 = 10 N
108 27q
e e
x y z x y zF E
5 5
0 0
ˆ ˆ ˆ ˆ2 2 2 4 6 ˆ ˆ ˆ1 4 210 10 V/m
4 27 216 108 e e
x y z x x y zE
• When we have a continuous charge density, each increment of
charge contributes
• Integrating over a complete volume, we obtain:
3-3.2 Electric Field due to a Charge Distributions
vector from differential
volume to point
is the
dv P
R
Vdq dv
V
2 2ˆ ˆ , where
4 4
dvd qd
R R
e e
E R R
V
2
1 ˆ (volume distribution)4v v
dvd
R
e
E E R
• For surface and line distributions, these become
S
2
2
1 ˆ (surface distribution)4
1 ˆ ( line distribution)4
s
l
l
d s
R
d l
R
e
e
E R
E R
:Electric Field of a Ring of Charge1Example
• Determination of the electric field intensity 𝐸 at a
point 𝑃(0,0, ) along the axis of a ring at a distance
from its center.
𝑑𝑙 = 𝑏𝑑𝜙 = 𝜌𝑙 𝑏𝑑𝜙
𝑅1 = 𝑅1 = 𝑏2 + 2
𝑹 𝟏 =𝑅1
𝑅1=− 𝒓 𝑏 + 𝒛
𝑏2 + 2
• The electric field at 𝑃(0,0, ) due to charge segment 1 is :
𝑑𝐸1 =1
4𝜋𝜀0𝑹 𝜌𝑙𝑑𝑙
𝑅12=
𝜌𝑙𝑏
4𝜋𝜀0 (−𝒓 𝑏 + 𝒛 )
(𝑏2 + 2)3/2𝑑𝜙
• The electric field at 𝑃(0,0, ) due to charge segment 2
is :
𝑑𝐸2 =𝜌𝑙𝑏
4𝜋𝜀0 (𝒓 𝑏 + 𝒛 )
(𝑏2 + 2)3/2𝑑𝜙
• The sum of the electric field from the two contributions of
segments 1 and 2 is :
𝑑𝐸 = 𝑑𝐸1 + 𝑑𝐸2 = 𝒛 𝜌𝑙𝑏
2𝜋𝜀0
𝑑𝜙
(𝑏2 + 2)3/2
∴ 𝐸 = 𝒛 𝜌𝑙𝑏
2𝜋𝜀0(𝑏2 + 2)3/2
𝑑𝜙 = 𝒛 𝜌𝑙𝑏
2𝜀0(𝑏2 + 2)3/2
𝜋
0
• Since 𝜌𝑙 =𝑄
2𝜋𝑏 , where 𝑄 is the total charge,
𝑄 = 𝜌𝑙2𝜋𝑏 ⇒ 𝜌𝑙𝑏 =𝑄
2𝜋
∴ 𝐸 = 𝒛
4𝜋𝜀0(𝑏2 + 2)3/2
𝑄
: Electric Field of a Circular Disk of charge2Example
• A ring of radius 𝑟 and width 𝑑𝑟 has an area
𝑑𝑠 = 2𝜋𝑟 𝑑𝑟
and contains charge :
𝑑𝑞 = 𝜌𝑠𝑑𝑠
= 2𝜋𝜌𝑠𝑟𝑑𝑟
• Since for a ring :
𝑑𝐸 = 𝒛
4𝜋𝜀0(𝑏2 + 2)3/2
𝑑𝑞
(replacing b with r)
𝑑𝐸 = 𝒛
4𝜋𝜀0 𝑟2 + 232
(2𝜋𝜌𝑠𝑟𝑑𝑟)
∴ 𝐸 = 𝒛 𝜌𝑠
2𝜀0
𝑟𝑑𝑟
(𝑟2 + 2)3/2
𝑎
0
= 𝒛 𝜌𝑠ℎ
2𝜀0 (1
2−2 )
1
(𝑟2+ℎ2)1/2 𝑟 = 0
𝑟 = 𝑎
𝐸 = 𝒛 𝜌𝑠
2𝜀0(1
−
1
𝑎2 + 2)
𝐸 = ± 𝒛 𝜌𝑠2𝜀0
(1 −
𝑎2 + 2)
Note that …
𝐸 is positive(+) when > 0 and is negative(-) when < 0
• For an infinite sheet of charge with 𝑎 = ∞
𝐸 = 𝒛 𝜌𝑠2𝜀0
Note that …
𝐸 =
𝒛 𝜌𝑠2𝜀0
𝑧 > 0
−𝒛 𝜌𝑠2𝜀0
𝑧 < 0
• Use coulomb's law to obtain an expression for 𝐸 in a
line of charge with uniform density 𝜌𝑙 extends between
𝑦 = −𝐿 2 𝑎𝑛𝑑 𝑦 = 𝐿 2 along the y − axis at a point
𝑃 located on the 𝑥- axis as shown in the figure.
Example 3 Electric field of an infinite line of charge
𝒓 =𝒙 𝑥 − 𝒚 𝑦
𝑥2 + 𝑦2
𝐸 = 𝑘(𝑑𝑞)
𝑟2𝑙
𝒓
𝐸 = 𝑘𝜌𝑙𝑑𝑦
(𝑥2+𝑦2)
𝐿/2
−𝐿/2
𝒙 𝑥 − 𝒚 𝑦
𝑥2 + 𝑦2
*** from Coulomb's law,
,Solution
𝐸 = 𝑘𝜌𝑙𝑑𝑦
(𝑥2+𝑦2)
𝐿/2
−𝐿/2
𝒙 𝑥 − 𝒚 𝑦
𝑥2 + 𝑦2
𝐸 = 𝑘𝜌𝑙(𝒙 𝑥 𝑑𝑦
𝑥2 + 𝑦232
𝑙2
−𝑙2
− 𝒚 𝑦𝑑𝑦
(𝑥2 + 𝑦2)3/2)
𝑙/2
−𝑙/2
𝑦 components cancel each
other due to symmetry.
𝑘 = 1/4𝜋𝜀0 , 𝑦 = 𝑥 tan𝜃 , 𝑑𝑦 = 𝑥 sec2 𝜃𝑑𝜃 = 𝑥
cos2 𝜃 𝑑𝜃
𝑑𝑙
𝑑𝜃=
𝑥
cos2 𝜃⇒ 𝑑𝑦 =
𝑥
cos2 𝜃𝑑𝜃
𝐸 = 𝒙 𝜌𝑙
2𝜋𝜀0𝑥
4Example
𝜌𝑠
solution
𝜌𝑠
𝜌𝑠
𝜌𝑠
Exercise
Solution 𝜌𝑙 = 𝜌𝑠𝑑𝑦
𝐸 =𝜌𝑙
2𝜋𝜀0𝑅
𝑑𝐸 = 𝒛 2𝜌𝑠𝑑𝑦 cos 𝜃
2𝜋𝜀0𝑅
𝐸 = 𝑑𝐸 = 𝒛 𝜌𝑠𝜋𝜀0
cos 𝜃
𝑅 𝑑𝑦
𝑑/2
0
𝑑/2
0
cos 𝜃 =
𝑅 tan 𝜃 =
𝑦
∴ 𝑦 = tan 𝜃 𝑑𝑦 = sec2 𝜃 𝑑𝜃
𝐸 = 𝒛 𝜌𝑠𝜋𝜀0
cos2 𝜃
𝜃0
0
. sec2 𝜃 𝑑𝜃
= 𝒛 𝜌𝑠𝜋𝜀0
𝜃0
*** For an infinitely wide sheet,𝜃0 =𝜋
2, 𝑡𝑒𝑛
𝐸 = 𝒛 𝜌𝑠
2𝜀0 (when 𝑑 is infinite)
3-4 Gauss’s Law
• We have following differential form of Gauss’s law :
𝛻. 𝐷 = 𝜌𝑣 (Gauss's law)
• It can be converted and expressed in integral form by multiplying both sides by 𝑑𝑣:
𝛻. 𝐷𝑑𝑣𝑣
= 𝜌𝑣𝑑𝑣 = 𝑄𝑣
,where 𝑄 is the total charge enclosed in volume 𝑣.
• The divergence theorem:
𝛻. 𝐸 𝑑𝑣𝑣
= 𝐸. 𝑑𝑠 𝑠
States that the volume integral of the divergence
of any vector over a volume 𝑣 is equal to the
total outward flux of that vector through the
surface 𝑆 enclosing 𝑣.
• For the vector 𝐷 :
𝛻.𝐷 𝑑𝑣𝑣
= 𝐷 . 𝑑𝑠 𝑠
• Comparing all these equations, we have ;
𝐷 . 𝑑𝑠 𝑠
= 𝑄 (𝐺𝑎𝑢𝑠𝑠′𝑠 𝑙𝑎𝑤)
• As shown in the figure, each differential surface
element 𝑑𝑠 , 𝐷 . 𝑑𝑠 is the electric field flux flowing
out- wardly through 𝑑𝑠 ,and the total flux through
the surface S is equal to the enclosed charge Q.
• The surface 𝑆 is called a Gaussian surface.
• If the dimensions of a very small volume ∆𝑣
containing a total charge 𝑞 are much smaller than
the distance from ∆𝑣 to the point at which the
electric flux density 𝐷 is to be evaluated, then 𝑞
may be regarded as a “point charge”.
• At any point on the surface defined by positive
vector 𝑅 .
𝐷 = 𝑹 𝐷𝑅
𝑑𝑠 = 𝑹 𝑑𝑠
• Applying Gauss’s law ,
𝐷 . 𝑑𝑠 𝑠
= 𝑹 𝐷𝑅 . 𝑹 𝑑𝑠𝑠
= 𝐷𝑅𝑑𝑠𝑠= 𝐷𝑅 4𝜋𝑅2 = 𝑞
𝐷𝑅 =𝑞
4𝜋𝑅2
∴ 𝐷 = 𝑹 𝑞
4𝜋𝑅2
• The direction of 𝐷 must be radially outward along the unit vector
𝐑 , and 𝐷𝑅 , 𝑡𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝐷,must be the same at all points on the Gaussian surface S.
=1
• Since 𝐷 = 𝜀𝐸 ⇒ 𝐸 =𝐷
𝜀0 ( in free space)
∴ 𝐸 = 𝑹 𝑞
4𝜋𝜀𝑅2 (𝑉 𝑚 )
• Although Coulomb’s law can be used to find 𝐸 for any
specified distribution of charge, Gauss's law is easier to
apply than Coulomb’s law, but its utility is limited to
symmetrical charge distributions.
(in a medium with permittivity 𝜀)
Example 1
• Use Gauss’s law to obtain an expression for electric field 𝐸 due to
an infinite line of charge with uniform charge density 𝜌𝑙 along z-
axis.
• By symmetry: 𝐸 = 𝒓 𝐸𝑟
𝐷. 𝑑𝑠 =0 on the top and bottom surfaces.
𝐸𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 on lateral surface= 𝐸
𝑄 = 𝐷 . 𝑑𝑠 𝑠
= 𝜀0𝐸 . 𝑑𝑠 𝑠
= 𝜀0 𝐸𝑟 .2𝜋𝑟
∴ 𝐸𝑟 =𝑄
2𝜋𝜀0𝑟 ℎ =
𝜌𝑙
2𝜋𝜀0𝑟
∴ 𝐸 = 𝒓 𝜌𝑙
2𝜋𝜀0𝑟
= 𝜌𝑙
*** Construct an imaginary
Gaussian cylinder of radius r and
height h.
Solution:
Which is the same as
integrating charge distribution.
(infinite line of charge)
2 Example
Determine the electric field 𝐸 at a distance 𝑧 from an
infinitely large thin charged sheet with
a uniform surface charge density 𝜌𝑠 .
solution
• The direction of 𝐸 is along z-axis.
Therefore, choose Gaussian surface
to be a cylinder whose axis is aligned with
the z-axis.
𝑄 = 𝐷 . 𝑑𝑠 𝑠
= 𝜀0𝐸 . 𝑑𝑠 𝑠
𝐸 . 𝑑𝑠 = 0 on lateral surface. (𝐸 ⊥ 𝑑𝑠 ) (no flux).
On the top and bottom: 𝐸 ∥ 𝑑𝑠 , 𝐸 is constant (same flux).
∴ 𝑄 = 𝜀0 𝐸 . 𝑑𝑠 𝑠
= 𝜀0 𝐸𝐴𝑡𝑜𝑝 + 𝐸𝐴𝑏𝑜𝑡𝑡𝑜𝑚 = 2𝜀0𝐸𝐴
∴ 𝐸 = 𝒛 𝑄
2𝜀0𝐴= 𝒛
𝜌𝑠
2𝜀0 (infinite sheet of charge)
*** The result is worth remembering:
Infinite sheet of charge produces uniform 𝐸 field of 𝜌𝑠 2𝜀0 above and below.
Which is the same as
integrating charge distribution.
Exercise
solution
3-5 Electric Scalar Potential
• In electric circuits we work with voltages and currents. The
voltage is the amount of work(or potential)required to move a
unit charge from one point to the other.
Voltage=Voltage potential=Electric Potential
• The voltage is a result of the existence of the electric field(𝐸) .
In this section, we want to find the relation between voltage V
and electric field 𝐸.
•
note that : Voltage is a scalar quantity….i.e. there are no vectors involved in
the calculations.
3-5.1 Electric Potential as a Function of Electric Field
• Consider a positive charge q in a uniform electric
field 𝐸 = −𝒚 𝐸, parallel to the – 𝑦 − direction.
• The presence of the field 𝐸 exerts a force 𝐹 𝑒 = 𝑞𝐸
(in the negative-y direction).
• Moving the charge along the positive y-
direction(against the force 𝐹 𝑒) requires applying an
external force 𝐹 𝑒𝑥𝑡 to counteract 𝐹 𝑒,which requires
the expenditure of energy.
• To move 𝑞 at constant speed , the net force acting on the charge must be zero.
𝐹 𝑒𝑥𝑡 + 𝐹 𝑒 = 0
∴ 𝐹 𝑒𝑥𝑡 = −𝐹 𝑒 = −𝑞𝐸
• The work done in moving the charge a vector
differential distance 𝑑𝑙 under the influence of
𝐹 𝑒𝑥𝑡 is :
𝑑𝑊 = 𝐹 𝑒𝑥𝑡 . 𝑑𝑙 = −𝑞𝐸. 𝑑𝑙 (J)
• If the charge is moved a distance 𝑑𝑦 along
𝒚 ,thus
𝑑𝑊 = −𝑞 −𝒚 𝐸 . 𝒚 𝑑𝑦 = 𝑞𝐸𝑑𝑦
• The differential electric potential energy 𝑑𝑊 per unit
charge is called the differential electric potential 𝑑𝑉.
𝑑𝑉 =𝑑𝑊
𝑞=1
𝑞−𝑞𝐸. 𝑑𝑙 = − 𝐸 . 𝑑𝑙 (
𝐽
𝐶 𝑜𝑟 𝑉)
• The potential difference between any two points 𝑃1 and 𝑃2 is
obtained by integrating 𝑑𝑉 along any bath between them .
𝑑𝑉𝑃2
𝑃1
= − 𝐸. 𝑑𝑙 𝑃2
𝑃1
∴ 𝑉21= 𝑉2 − 𝑉1 = − 𝐸. 𝑑𝑙 𝑃2
𝑃1
,where 𝑉1 and 𝑉2 are the electric potentials of points
𝑃1 and 𝑃2 respectively.
• The result of the line integral on the right- hand side of Eq 𝑉21 should be independent of the path taken between points 𝑃1 and 𝑃2.
• This requirement is mandated by the law of conservation of energy.
• The voltage difference between two nodes in an electric circuit has the same value regardless of the path taken between the nodes.
• Also Kirchhoff’s law of voltage states that
“the net voltage drop around a closed loop is
zero”.
• This implies that :
𝐸 . 𝑑𝑙 = 0 (𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐𝑠)𝑐
• The vector field whose line integral along any
closed path is zero is called conservative or
irrotational field.
• If 𝐸 is time- varying function, it is no longer
conservative, and its line integral along a
closed path is not necessary equal to zero.
• The conservative property of the electrostatic
field can also be deduced from Maxwell’s
second equation,
If 𝜕
𝜕𝑡= 0, then
𝛻 × 𝐸 = 0
• This Eq is the differential-form equivalent of
the Eq : 𝐸 . 𝑑𝑙 = 0𝑐
• If we apply stocke’s theorem to convert the
surface integral into a line integral, we obtain:
𝛻 × 𝐸 . 𝑑𝑠 = 𝐸 . 𝑑𝑙 = 0𝐶𝑆
,where 𝐶 is a closed contour surrounding 𝑆.
• The electric potential 𝑉 at any point 𝑃 is given
by:
𝑉 = − 𝐸 . 𝑑𝑙 (𝑉)
𝑃
∞
,Where the reference potential point 𝑃1 is chosen to be at
infinity; i.e. 𝑉1 = 𝑧𝑒𝑟𝑜 when 𝑃1 is at infinity.
3-5.2 Electric Potential due to Point Charge
• Since ; 𝐸 = 𝑹 𝑞
4𝜋𝜀𝑅2 (
𝑉
𝑚)
and 𝑑𝑙 = 𝑹 𝑑𝑅
∴ 𝑉 = − 𝑹 𝑞
4𝜋𝜀𝑅2. 𝑹 𝑑𝑅
𝑅
∞
𝑉 = − 𝑞
4𝜋𝜀𝑅2𝑑𝑅
𝑅
∞ =
𝑞
4𝜋𝜀𝑅
𝑅
∞
𝑉 =𝑞
4𝜋𝜀𝑅 (𝑉)
• If the charge 𝑞 is located at another position rather than the origin, and specified by the position vector
𝑅1 then,
𝑉 𝑅 =𝑞
4𝜋𝜀 𝑅−𝑅1
,where 𝑅 − 𝑅1 is the distance between the
observation point and the location of the charge 𝑞.
• For N point charges 𝑞1, 𝑞2, ……,𝑞𝑁 having
position vectors 𝑅1, 𝑅2,….., 𝑅𝑁 the electric
potential is :
𝑉 𝑅 =1
4𝜋𝜀
𝑞𝑖
𝑅 − 𝑅𝑖
𝑁
𝑖=1
(𝑉)
3-5.3 Electric potential due to Continuous Distribution
• In the final equation of electric potential due to
point charges if we :
1- replace
𝑞𝑖 𝑤𝑖𝑡 𝜌𝑣𝑑𝑣, 𝜌𝑠𝑑𝑠, and 𝜌𝑙𝑑𝑙 respectively and
2- convert the summation into an integration and
3- define 𝑅 = 𝑅 − 𝑅𝑖 as the distance between the
integration point and the observation point we get :
𝑉 𝑅 =1
4𝜋𝜀 𝜌𝑣𝑅
𝑣
𝑑𝑣 (Volume distribution)
𝑉 𝑅 =1
4𝜋𝜀 𝜌𝑠𝑅
𝑠
𝑑𝑠 (Surface distribution)
𝑉 𝑅 =1
4𝜋𝜀 𝜌𝑙𝑅
𝑙
𝑑𝑙 (Line distribution)
• From previous lectures we have:
𝑑𝑉 = −𝐸 . 𝑑𝑙
• Also we have for a scalar function 𝑉
𝑑𝑉 = 𝛻𝑉 . 𝑑𝑙
• Comparing the last two equations, we get
𝐸 = −𝛻𝑉
3-5.4 Electric Field as a Function of Electric Potential
Why minus?!
We know from the previous lectures that the gradient
operator is a differential operator which operates on a scalar
function to yield :
1- the maximum increase per unit distance.
2- the direction of maximum increase.
*** Since the electric field always points in the direction of
decreasing potential, the electric field is the negative of the
gradient of 𝑉.
1 Example
An electric dipole consists of two point charges of equal
magnitude and opposite polarity, separated by a small distance 𝑑
as shown in the figure . Determine
(A) the electric potential (𝑉) and
(B) the magnitude of the electric field 𝐸 at any point in free space
given that 𝑃 is at distance 𝑅 ≫ 𝑑.
Solution
,Where 𝑝 = 𝑞𝑑 is called the dipole moment of the
electric dipole.
(b) The far-field electric field produced by the electric dipole is found
by taking the gradient of the potential in spherical coordinates.
𝐸 = −𝛻𝑉 = − 𝑹 𝜕𝑉
𝜕𝑅+ 𝜽
1
𝑅
𝜕𝑉
𝜕𝜃+ 𝝓
1
𝑅 sin 𝜃
𝜕𝑉
𝜕𝜙
𝑉 =𝑞𝑑 cos 𝜃
4𝜋𝜀0𝑅2
= − 𝑞𝑑
4𝜋𝜀0𝑹 cos 𝜃
𝜕
𝜕𝑅
1
𝑅2+ 𝜽
1
𝑅3
𝜕
𝜕𝜃(cos 𝜃)
= − 𝑞𝑑
4𝜋𝜀0𝑹 cos 𝜃 −
2
𝑅3+ 𝜽
1
𝑅3 (− sin 𝜃)
∴ 𝐸 = 𝑞𝑑
4𝜋𝜀0𝑅3 𝑹 2 cos 𝜃 + 𝜽 sin 𝜃 𝑉 𝑚 , 𝑅 ≫ 𝑑
2Example
A uniformly charged disk has radius a and
surface charge density 𝜌𝑠. Find :
(A) the electric potential and
(B) the magnitude of the electric field along the
perpendicular central axis of the disk.
solution
𝑑𝑠 = 2𝜋𝑟 𝑑𝑟. 𝑑𝑞 = 𝜌𝑠 𝑑𝑠 = 𝜌𝑠 2𝜋𝑟 𝑑𝑟.
𝑉 =1
4𝜋𝜀0
𝜌𝑠2𝜋𝑟𝑑𝑟
𝑥2 + 𝑟2
𝑎
0
𝑉 =𝜌𝑠𝜋
4𝜋𝜀0
2𝑟𝑑𝑟
𝑥2 + 𝑟2
𝑎
0
𝑉 =𝜌𝑠𝜋
4𝜋𝜀0 2𝑟𝑑𝑟(𝑥2 + 𝑟2)−1/2𝑎
0
𝑉 =𝜌𝑠2𝜀0
𝑥2 + 𝑎2 1/2 − 𝑥
𝐸𝑥 = −𝛻𝑉 = 𝒙 𝜕𝑉
𝜕𝑥= 𝒙
𝜌𝑠
2𝜀01 −
𝑥
𝑥2+𝑎2 𝑉 𝑚 , 𝑥 > 0
3-5.5 Poisson’s Equation
• Since 𝐷 = 𝜀𝐸 , the differential form of Gauss’s law:
𝛻 . 𝐷 = 𝜌𝑣
may be written as
𝛻 . 𝐸 =𝜌𝑣
𝜀
• Since 𝐸 = −𝛻𝑉 from previous lectures ,then
𝛻 . 𝛻𝑉 = − 𝜌𝑣𝜀
• From the definition of Laplacian of a scalar function :
𝛻2𝑉 =𝜕2𝑉
𝜕𝑥2 +
𝜕2𝑉
𝜕𝑦2 +
𝜕2𝑉
𝜕𝑧2
∴
𝛻2𝑉 = − 𝜌𝑣
𝜀 (Poisson’s equation)
• For a volume 𝑣′ containing a volume charge
density distribution 𝜌𝑣,the solution for 𝑉 was
expressed before as:
𝑉 =1
4𝜋𝜀
𝜌𝑣
𝑅′ 𝑑𝑣′
𝑣′
which satisfies the Poisson's equation. If the
medium under consideration contains no free
charges, then:
𝛻2𝑉 = 0 (𝐿𝑎𝑝𝑙𝑎𝑐𝑒′𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛)
*** Poisson’s and Laplace’s equations are useful to find the electrostatic
potential 𝑉 in a region whose boundaries 𝑉 is known.(e.g. region between the
plates of a capacitor with a specified voltage difference across it).