Chapter two: Electrostatics Coulomb's law: The force on a test ...

Electromagnetic <strong>The</strong>orem 

(Dr. Omed Ghareb Abdullah) University of Sulaimani –College of Science – Physics Department 

<strong>Chapter</strong> <strong>two</strong>: <strong>Electrostatics</strong> 

Coulomb’s <strong>law</strong>: 

<strong>The</strong> <strong>force</strong> on a test charge due to a single point charge is proportional to the charges 

and inversely proportional to the square of the separation distance. 

1 

4 ̂ 

where is permittivity of free space 

8.8510 

 

 

. 

 

 

If we have several point charges , , , ⋯ , 

at distances , , , ⋯ , from test charge . 

<strong>The</strong> total <strong>force</strong> on a is: 

 

where 

⋯ 

1 

4 

̂ 

 

̂ 

 

̂ ⋯ 

 

 

 

4 

̂ 

 

̂ 

 

̂ ⋯ 

 

 

 

1 

4 

̂ 

 

 

: is called the electric field of the source charges. 

 

 

 

 

 

 

 

 

 

76



Problem: 

Find the electric field a distance above the midpoint between <strong>two</strong> equal charges, , a 

distance apart. 

Repeat part , only make the right‐hand charge – instead of . 

 

 

 

 

 

 

/ 

/ 

 

 

Horizontal components cancel. 

Vertical field is: 

1 

4 

2 

cos 

when 

; cos 

1 2 

4 / 

≫ ⇒ 1 2 

4 

It looks like a single charge 2. 

If 0 ⇒ 0 

 

 

 

 

 

 

 

/ 

/ 

 

 

1 

4 

2 

sin 

; sin / 

 

77



when 

1 

4 

 

/ 

≫ . . ⟶ 0 ⇒ 0 

If 0 ⇒ 

 

 

 

Problem: 

Charge 4 is located at 1,1,0 and charge is located at 0,0,4. What 

should be so that at 0,2,0 has no ‐component? 

 

 

 

 

 

 

 

 

 

 

1 

4 

 

 

1 

4 

4 

√2 1 

2 

1 

4 

1 

4 √20 1 

4 20 

cos 1 

2 

1 

√2 

cos 1 

4 20 2 

√20 

0 

 

2 √2 

10 √20 0 

20√10 10√40 63.2455 . 

78



Continuous Charge Distributions: 

If the charge is distributed continuously over some region, the sum becomes an integral: 

1 

4 ̂ 

Thus, the electric field of a charge is: 

1 

4 ̂ 

For a surface charge: 

1 

4 ̂ 

and, for a volume charge: 

1 

4 ̂ 

where: 

: Charge per unit length, : Charge per unit area, : Charge per unit volume. 

: an element of length along the line, : an element of area on the surface, and 

: an element of volume. 

Example: 

Find the electric field a distance above the midpoint of a straight line segment of length 2 

which carries a uniform line charge . 

 

 

 

 

 

 

 

 

Due to symmetric horizontal components = 0. 

1 

4 

 

̂ 

 

1 

2 

 

4 cos 

 

79



cos 

; ; : 0 → 

1 2 

 

4 

 

 

 

/ 

 

2 

 

4 / 

 

2 

 

 

4 √ 

1 2 

4 √ 

⋙ For points far from the line ≫ : 

≅ 1 2 

4 

It looks like a point charge 2. 

⋙ For points ≫ : 

≅ 1 

4 

2 

 

Notes: 

Let: 

 

 

 

 

/ 1 / 

 

tan 

⇒ tan ⇒ sec 

 

 

 

sec 

/ 1 / 

1tan sec 

 

 

 

sec 

/ sec 

 

/ sec 

sin 

 

 

/ 

 

sin 

√ 

 

 

/ 

√ 

cos 

 

 

80



Problem: 

Find the electric field a distance above one end of a straight line segment of length , which 

carries a uniform line charge . Check that your formula is consistent with what you would 

expect for the case ≫. 

 

 

 

 

 

 

 

1 

4 

 

̂ 

 

1 

4 cos 

 

cos 

; ; : 0 → 

1 

 

4 

 

 

/ 

 

 

 

 

 

4 √ 

1 

4 √ 

1 

 

4 sin 

 

 

4 

 

 

 

 

 

 

 

4 

 

1 

/ 

 

 

√ 

 

1 4 1 

√ 

81



1 

4 

 

1 

 

 

√ 

√ 

⋙ For points far from the line ≫ ; you expect it to look point charge . 

≅ 1 

4 

Problem: 

Find the electric field a distance above the center of a square loop (side ) carrying 

uniform line charge . 

 

 

 

 

/ 

 

 

1 

4 

 

̂ 

1 

 

4 cos 

1 

 

 

 

4 

 

 

For one side: 

/ 

 

4 

 

 

 

/ 

 

 

 

 

2 

 

4 

 

 

 

 

2 

 

/2 

4 

 

 

 

 

/ 

 

 

 

82



<strong>The</strong>re are four sides: 

 

 

4 

4 

4 

 

 

 

 

 

 

 

 

 

 

Problem: 

Find the electric field a distance above the center of a circular loop of radius , which 

carries a uniform line charge . 

 

 

 

 

 

Horizontal components cancel; 

Here: 

1 

4 

 

̂ 

 

 

 

4 cos 

 

 

 

4 / 

∶ 

cos 

∶ 

 

 

 

1 

4 / 

1 

4 

 

2 

/ 

83



Problem: 

Find the electric field a distance above the center of a flat circular disk of radius , which 

carries a uniform surface charge . 

 

 

 

 

 

 

 

1 

4 

 

 

1 2 

 

4 cos 

 

2 

 

4 √ 

2 

 

4 / 

2 

4 2 

√ 

 

 

2 

4 

1 1 

√ 

Flux and Gauss’s <strong>law</strong>: 

<strong>The</strong> electric field of a single point charge is: 

1 

4 ̂ 

<strong>The</strong> flux of through a surface of radius is: 

∙ 1 

4 ̂∙ sin ̂ 

 

sin 

22 

4 4 

∙ 

This is a Gauss’s <strong>law</strong>; which suggests that the flux through any closed surface is a measure of 

the total charge inside. 

84



As it stands, Gauss’s <strong>law</strong> is an integral equation, but we can readily turn it into a differential 

one, by applying the divergence theorem. 

Rewriting in terms of charge density . 

So Gauss’s <strong>law</strong> become: 

or 

∙ ∙ dτ 

 

∙ dτ 

 

Gauss’s <strong>law</strong> in differential form. 

∙ 

Problem: 

Suppose the electric field in some region is found to be , in spherical 

coordinates ( is constant). 

Find the charge density . 

Find the total charge contained in a sphere of radius , centered at the origin. 

a) 

∙ 

⇒ ∙ 

b) from Gauss’s <strong>law</strong>: 

∙ 1 

1 

5 

⇒ 

5 

∙ 

∙ ∙ sin 

or: 

4 4 

 

5 sin 

 

85



 

 

5 sin 

 

 

 

5 

5 22 4 

Example: 

Find the field outside a uniformly charged solid sphere of radius and total charge . 

Drawing a spherical surface at radius . 

 

 

 

 

∙ 

 

∙ sin 

22 

1 

4 

Problem: 

Use Gauss’s <strong>law</strong> to find the electric field inside and outside a spherical shell of radius . 

Which carries a uniform surface charge density . 

 

 

 

∙ 

Inside: 0 

∙ 4 0 ⇒ 0 

 

86



Outside: 4 

∙ 4 4 

⇒ 

 

Problem: 

Use Gauss’s <strong>law</strong> to find the electric field inside a uniformly charged sphere (charge density ) 

 

 

∙ 

∙ 4 

sin 

∴ 4 1 

 

 

3 

⇒ 

3 

 

Problem: 

Find the electric field a distance from an infinitely long straight wire, which carries a 

uniform line charge . 

 

 

∙ 

∙ 2 

 

 

2 

 

∴ 2 

 

⇒ 

2 

87



Problem: 

An infinite plate carries a uniform 

surface charge . Find 

its electric field. 

 

∙ 

In this case ; from the top and bottom surface yield 

∙ 2 

 

2 

 

∴ 2 

⇒ 

 

 

2 

Results: 

⊚ <strong>The</strong> electric field of a sphere falls off like 1/ 

⊚ <strong>The</strong> electric field of an infinite 

line falls off like 1/ 

⊚ <strong>The</strong> electric field of an infinite 

plane does 

not falls offf at all. 

Divergence and Curl of 

∙ 

 

4 

1 

 

4 ̂ 

∙ ̂ 

 

 

 

̂ 

4 

 

⇒ 

⇒ 

∙ 

0 

Problem: 

Find the electric field 

inside a sphere which carries a charge density proportional to the 

distance from the origin, , for some constant .



 

 

∙ 

∙ 4 

sin 

 

∴ 4 

 

22 

1 

4 

⇒ 1 

4 

 

Problem: 

A hollow spherical shell carries a charge density, , in the region . 

Find the electric field in the three regions: 

 

(i) 

∙ 

(ii) 

0 ⇒ 0 

 

 

 

∙ 4 

sin sin 

 

22 4 

∴ 4 

 

 

4 

 

 

 

⇒ 

 

 

 

 

 

89



(iii) 

 

 

 

∙ 4 

 

 

 

sin sin 

22 4 

∴ 4 

 

 

4 

 

 

 

⇒ 

 

 

 

Example: 

A long cylinder carries a charge density that is proportional to the distance from the axis 

for some constant . Find the electric field inside this cylinder. 

 

 

∙ 

∙ 2 

 

 

 

 

 

 

 

 

3 2 2 3 

∴ 2 1 

2 

3 

1 

3 

⇒ 1 

3 

 

90

Electromagnetic <strong>The</strong>orem (Dr. Omed Ghareb Abdullah) University of Sulaimani –College of Science – Physics Department 

Example: 

Two infinite parallel planes carry 

equal but opposite uniform charge 

density . Find the 

field in each of the three regions: 

to the 

left of both. 

between them. 

to the right of both. 

 

 

 

 

 

 

 

 

 

 

 

<strong>The</strong> left plate produce a field 

which points away from it. 

 

<strong>The</strong> right plate produce a field which points toward 

it. 

 

 

0 

 

2 2 

 

0 

Problem: 

A charge sits at the 

back corner of a cube, as shown in Figure. What is the flux 

of through 

the shades side? 

Think of this cube as one of (8) surrounding the charge. Each of the (24) squaress which make 

up the surface of this 

large cube gets the same flux as every other one, so: 

But; 

 

 

∙ 

1 

24 

∙



 

 

∴ ∙ 

 

∙ 

 

 

 

24 

Problem: 

One of these is an impossible electrostatic field, which one? 

 

2 3 

 

2 2 

Here is a constant. 

 

 

 

 

02 03 0 0 

 

2 3 

So is an impossible electrostatic field. 

 

 

 

 

2 2 

 

00 2 2 0 

2 2 

So is a possible electrostatic field. 

Problem: 

If the electric field in some region is given by the expression: 

sincos 

 

where & are constant, what is the charge density? 

1 r A 1 

rsinθ 

∙ 

⇒ ∙ 

1 ∂ A 

r ∂r r r 1 ∂ 

rsinθ∂φ sincos 

 

sin 

 

sin ⇒ 

AB sin 

r 92



Problem: 

A long 

coaxial cable carries a uniform volume charge density on 

the inner cylinder 

(radius ), and a uniform surface 

charge density on the outer cylindrical shell (radius ); This 

surface charge is negative and of 

just the right magnitude so that the cable as a whole is 

electrically neutral. Find the electric field in each of the three regions: 

a) Inside the inner cylinder . 

b) Between the cylinders . 

a) Outside the cable . 

a)) 

 

∙ 

∙ 2 

 

∴ 2 

1 

 

b)) 

 

 

2 

⇒ 

 

 

2 

 

∙ 2 

 

∴ 2 

1 

 

c)) 

 

 

2 

 

⇒ 

 

 

2 

 

∙ 2 

0 

0



Electric Potential: 

Electric potential is define as: 

 

. 

 

: standard reference point. 

<strong>The</strong> potential difference between <strong>two</strong> point & is: 

 

 

. 

. 

 

 

 

 

 

. . . 

 

 

 

From, the 

fundamental theorem for gradients: 

 

. 

 

 

 

 

. . 

 

 

 

<strong>The</strong> electric field is the gradient of a scalar potential. 

Example: 

Find the potential inside and outside a spherical shell of radius , which carries a uniform 

surface charge. Set the referencee point at infinity. 

1 

 

4 ̂ 

For points outside the sphere : 

 

 

1 

 

. 4 1 

 

4 1 

4 

∞ 

∞ 

∞ 

To find the potential inside the sphere , we must break the integral into <strong>two</strong> sections: 

 

 

 

1 

4 1 

0 

4 1 

 

4 

∞ 

∞



Problem: 

Find the potential inside and outside a uniformly charged sphere whose radius is , and 

whose total charge is 

. Use infinity as your reference point. 

Outside the sphere : 

Inside the 

sphere : 

For : 

For : 

1 

4 

∞ 

 

 

. 

1 

4 ̂ 

1 

4 ̂ 

 

1 

 

4 

 

1 

4 

∞ 

 

 

1 4 1 

1 

4 2 3 

 

∞ 

 

1 

4 

 

 

 

 

1 

4 

∞ 

 

 

 

Problem: 

Find the potential a distance from an infinitely long straight wire that carries a uniform line 

charge . . 

<strong>The</strong> electric field of nfinite long straight wire 

is: 

1 2 

4 

In this case we cannot set the reference point at ∞. 

Lets set it 

at : 

 

1 2 

4 1 

2 ln 

4



Poisson’s Equation & Laplace’s Equation: 

We have: 

and 

∙ 

∴ ∙ ⇒ 

This is known as Poisson’s equation. 

In region where there is no charge, so that 0, Poisson’s equation reduces to Laplace’s 

equation: 0 

<strong>The</strong> potential of a localized charge distribution: 

<strong>The</strong> potential of a point charge at the origin is: 

. 1 

4 

1 

4 

<strong>The</strong> potential of collection of charges is: 

1 

4 

For a continuous distribution: 

1 

4 

For a volume charge, it’s: 

1 4 

Problem: 

Find the potential at a distance above the center of: 

(1) a straight line of length 2, which carries a uniform line charge . 

(2) a flat circular disk of radius , which carries a uniform surface charge . 

(1) 

96



 

 

2 

 

 

or: 

 

1 

4 

 

1 

4 

 

 

ln √ 

 

4 

 

 

 

 

 

√ 

 

4 

ln √ 

√ 

 

sinh 

4 

2 sinh 

4 

(2) 

 

 

 

 

 

1 

4 1 

4 

1 

4 

2 

 

 

 

 

2 

√ 

 

 

2 

 

Problem: 

A conical surface carries a uniform surface charge . <strong>The</strong> height of the cone is , as is the 

radius of the top. Find the potential difference between positions (the vertex) and (the 

center of the top). 

 

 

 

’ 

 

 

97



where; √2 

√ 

1 2 

4 

√ 

2 /√2 

4 

 

 

 

√ 

1 2 

4 ′ 

√ 

2 

4 

1 

√2 

 

 

 

 

2 

√2 

√2 

2 

 

 

√2 

 

√2 

2√2 √2 ln2 √2 2√2 

 

 

2√2 √2 ln2 2√2√2 ln2 √2 

√2 

 

 

2√2 

 

 

ln2 √2 ln2 √2 

√2 

ln 2√2 2 √2 

ln ln1 √2 

4 2√2 4 2 2 

∴ 

2 

1 ln1 √2 

√ 

<strong>The</strong> three fundamental quantities of electrostatics are: , , & : 

 

 

 

∙ 

 

98



Electrostatic Boundary Conditions: 

Suppose we draw a wafer‐thin Gaussian pillbox, extending just barely over the edge in each 

direction. 

Gauss’s <strong>law</strong> state that: 

∙ 

 

Where is the area of the pillbox lid in the limit as the thickness goes to zero. 

 

 

 

 

 

E 

 

a <strong>force</strong> oppo 

Work and 

Energy in <strong>Electrostatics</strong>: 

<strong>The</strong> work 

done to move a test charge q, you must exert osite to the electric field. 

 


is therefore: 

 

 

 

 

∙ ∙ 

∙ 

 

 

 

Notice that the answer is independent of the 

path you take from to . (the electric <strong>force</strong> 

are conservative). 

∴ 

 

<strong>The</strong> potential difference between points & is equal to the work per unit charge required 

to carry a particle from to . 

If we bring the charge in from far away and 

stick it at point , the work you must do is: 

∞ ⇒ 

<strong>The</strong> energy of a point charge distribution: 

How much work would it take to 

assemble an entire collection of point charges? 

Imagine bringing in the charges, one by one from far away:



 

 

 

 

 

 

<strong>The</strong> first charge takes no work. 

Work to bring is: 

Work to bring is: 

1 

4 

 

 

 

1 

4 

 

 

 

Similarly, the extra work to bring in will be: 

1 

4 

 

 

 

 

<strong>The</strong> total work necessary to assemble the first four charges, is: 

1 

4 

<strong>The</strong> general rule: 

 

 

 

 

 

 

 

 

1 

4 

 

 

 

 

 

 

 

 

 

 

 

 

<strong>The</strong> stipulation is just to remind you not to count the same pair twice, we can write: 

1 

8 

 

 

 

 

 

 

 

 

1 2 

 

 

1 

 

4 1 

2 

 

 

 

 

Problem: 

Three point charges 1 nC, 4 nC, and 3 nC, are located at 0 , 0 , 0, 0 , 0 , 1, and 1 , 0 , 0, 

respectively. Find the energy in the system. 

100



Solution: 

W 

WW W W 

W0q V q V V 

q 

Wq 

4πϵ 1 q q 

 

4πϵ 1 q 

4πϵ √2 

W 1 

4πϵ 

q q q q q q 

√2 

1 

4π8.85 10 4312 

√2 10 13.37 mJ 

<strong>The</strong> energy of a continuous charge distribution: 

For a volume charge density : 

1 2 

 

∙ 

⇒ ∙ 

Note: 

 

2 ∙ 

∙fA f ∙AA ∙f 

∙fA dτ f ∙A dτ A ∙f dτ 

 

∙fA dτ f A ∙da 

∴ 

f ∙A dτ A ∙f dτ f A ∙da 

To transfer the derivative from E to 

But: 

V E 

 

2 E ∙V dτ V E ∙da 

For large volume: 

 

2 E dτ V E ∙da 

 

2 

 

 

E dτ 

101



Problem: 

Find the energy stored in a uniformly charged solid sphere of radius and charge . 

Problem: 

 

2 

 

2 E dτ 

⇒ 1 

4 

 

̂ 

⇒ 1 

4 

 

̂ 

 

 

4 

 

1 

4 2 1 

5 

∞ 

4 1 

 

 

∞ 

1 

 

1 

4 2 1 

5 1 1 3 

4 5 

4 

Consider <strong>two</strong> concentric spherical shells, of radii & . Suppose the inner one carries a 

charge , and the other one a charge – (both of them uniformly distributed over the 

surface). Calculate the energy of this configuration. 

1 

4 

 

̂ 

 

2 E dτ 

, 

∴ 

2 

 

 

1 

4 

 

 

 

 

4 

1 8 

 

1 8 1 

Conductors: 

1‐ 0 inside a conductor. 

102



Putting a conductor in to an external field 

, this will drive any free positive charges to 

the right and negative ones to 

the left. Now these induced charges produce a field of their 

own , which is in 

the opposite direction 

to . So the field of induced charges tends to 

cancel 

off the original field, and the resultant field inside the conductor is precisely zero. 

2‐ 0 

inside a conductor, from 

Gauss’s <strong>law</strong> ∙ 

. If 0 ⇒ 0. 

 

3‐ Any net charge resides on the surface. 

4‐ A conductor is an equipotential. If & are <strong>two</strong> points within (or 

at the surface) a given 

 

conductor 

∙ 0, hence . 

 

5‐ is perpendicularr to the surface, just outside a conductor. 

Capacitos: 

<strong>The</strong> capacitance / 

Capacitance is a purely geometrical quantity, determined by the size, shapes and separation 

of the <strong>two</strong> conductors. 

For a parallel‐plat capacitor the surfaces of area , and held a distance apart: 

; E σ for plat ; and V Ed 

 

ϵ 

σ 

∴ d q d 

ϵ ϵ A 

& ⇒ ϵ A 

 

 


you must do to increase the charge 

by a small amount is: 

dq 

<strong>The</strong> total work necessary, to go from 0 to is: 

1 q 

⇒ 

1 2 C 

2 qV ⇒ 1 2 CV



Example: 

Find the capacitance of <strong>two</strong> concentric spherical metal shells, with radii & . 

1 

4 ̂ 

 

∙ 

 

 

1 4 

 

 

 

 

4 

1 1 

∴ 4 

 

 

More examples: 

Example(1): 

A charge distribution with spherical symmetry has density , determine everywhere and 

the energy stored in region . 

Outside the sphere: 

1 

4 

∭ ⟹ 

 

3 

Inside the sphere: 

∙ 

∙ 4 

∭ ⟹ 

∴ 4 

⟹ 

 

 

104



 

 

9 

 

 

 

 

 

 

 

 

 

4. 

 

18 5 2 

45 

 

 

. 

 

 

 

 

 

 

 

Example(2): 

A total charge of is uniformly distributed over a circular ring of radius . Find the potential 

at a distance above the center of the ring. 

Z 

r 

2 

z 

2 

Q dl 

dl d 

(0, ,z) 

V 

 

dl 

 

4 r 

2 

2 

0 

4 

Q 

Q dl 

(2 

) 

2 

Q 

V 

 

4 R 

o 

o 

 

z 

2 

d 

o 

2 

z 

2 

 

2 

o 

 

2 

z 

2 

r 

+ + + + + + 

+ 

+ + 

+ + 

+ + (,,0) 

Example (3): 

Determine , at 2,0, due to three charge distributions as follows; a uniform sheet at 

0 with 12 / , a uniform sheet at 4 with 12 / , and a 

uniform line at 6, 0 with 2 /. 

 

 

 

2 

 

2 

1 

4 

2 

 

, , 

 

 

12 

 

2 12 

 

2 

 

2 

2 4 

 

 

12.25 / 

Example (4): 

A charge distribution with spherical symmetry has density: 

105



 

0 

 

 

0 

Determine everywhere. 

a) For: : 

b) For: : 

 

4 sin 

 

 

 

 

4 

 

 

∴ 

4 

 

 

 

4 sin 

 

 

 

4 

 

 

 

∴ 

4 

Example (5): 

Giving that: 

2cos sin 

3 

 

in cylindrical coordinates, find the flux crossing the portion of the 0 plane defined 

by , 3/2 2. Assume flux positive in the direction. 

∙ 

1 3 

 

 

 

 

 

 

2cos sin 

3 

∙ 

/ 

sin 

/ 

 

1 3 cos / 

 

 

 

3 

Example (6): 

<strong>The</strong> charge is distributed along the z‐axis from to ∞ and to ∞ with a 

charge density of . Find on the ‐axis. 

 

 

4 

1 

4 

 

 

 

 

 

cos 

 

4 cos 

 

106



 

 

 

 

 

 

4 / 

/ 

 

 

 

 

 

 

4 √ 

√ 

 

 

 

4 

 

1 1 

√ √ 

 

 

4 2 1 

√ 

 

107

Chapter two: Electrostatics Coulomb's law: The force on a test ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?