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, 𝑡𝑕𝑒𝑛

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 , 𝑡𝑕𝑒 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).