ELECTRICAL FIELDS

I. Coulomb's law

F12 is the force exerted by the charge q1 on the charge q2
F12 is the force exerted by the charge q2 on the charge q1
We have F12 = F21.
The coulomb's law is written as :
F12 = F21 = F = (1/4πε0) q1 q2 /r2
F = (1/4πε0) q1 q2 /r2

II. The Electric field

If E is the field due to the charge q1
F12 = q2 E
Equating (1) and (2), we get:
E = (1/4πε0) q1  /r2
The expression of the electrical field E at the distance R from the charge Q is:
 E =  (1/4πε0) Q  ^r/ R2


The total Electrical field E due to n charges Qi distant ri 
from at the point P is 
E = (1/4πε0) Σ Qi ^ri/ ri2 ( i → n)

Recall that the electrical field E is a vector. It is directed 
allong with the unit vector ^r.

II.1. Due to two charges

If we want the total electrical field at the point P, we
can do the followings calculation:

E (total) = E1 (due to Q1) + E2 (due to Q2)

E1 = (1/4πε0) Q1 ^Q1P/ [Q1P]2
E2 = (1/4πε0) Q2 ^Q2P/ [Q2P]2

Recall that the ^r = r/|r|
Then:
R1 + Q1P = OP
Q1P = OP  - R1 = (4, -1) - (2,3) = (2, - 4)
|Q1P| = [(2)2 + (- 4)2]1/2 = [20]1/2
And:
^Q1P = Q1P/|Q1P|

R2 + Q2P = OP
Q2P = OP  - R2 = (4, -1) - (- 1, - 2) = (5, 1)
|Q2P| = [(5)2 + (1)2]1/2 = [26]1/2
And:
^Q2P = Q2P/|Q2P|

Thus:

E = (1/4πε0) ( Q1P . Q1/ |Q1P|3/2 + Q2P. Q2/ |Q2P|3/2)
E = (1/4πε0) ( Q1P . Q1/ (20)3/2 + Q2P . Q2/ (26)3/2) =
(1/4πε0) ( (2, - 4) . Q1/ (20)3/2 + (5, 1) . Q2/ (26)3/2 = 
(1/4πε0) ( 2 Q1/ (20)3/2 + 5 Q2/ (26)3/2, - 4 . Q1/ (20)3/2 + 1. Q2/ (26)3/2)

Or:
E = (1/4πε0) (2 Q1/(20)3/2 + 5 Q2/ (26)3/2) i +  (- 4 Q1/(20)3/2 + 1. Q2/ (26)3/2) j

Let: 
Q1 = - (20)3/2 µ C
Q2 = + (26)3/2 µ C
1/4πε0 = 9 x 10 9 N . m2/ C2 

Then E = 9 x 10 9[( - 2  + 5)i + (4 + 1)j ] . 1 x 10 - 6(N/C)
= 9 x 10 3[3i + 5j ](N/C) 
E = 9 x 10 3 [3  i + 5 j ] (N/C)

|E| = 9 x 10 3 [(3) 2 + ( 5) 2]1/2] (N/C) = 9 x 10 3 [34]1/2] (N/C)

|E| ≈ 50  x 10 3 N/C
This is the magnitude of the electrical field.
Or:
E = (1/4πε0) ( - Q1P  + Q2P)

II.2. Due to charged rod

Let's write dQ = λ dy = (Q/L) dy
Q is the charge of the whole rod; L is its length.
We have:
dEx = dE cos θ
dEy = dE sin θ
cos θ = b/R = b/ [y2 + b2 ]1/2
sin θ = y/R = y/ [y2 + b2 ]1/2

dEx = (1/4πε0) (dQ cos θ/R2) = (1/4πε0) (Q/L))(dy cos θ/R2)
dEy = (1/4πε0) (dQ sin θ/R2) = (1/4πε0) (Q/L) (dy sin θ/R2)

Then:
Ex = (1/4πε0) (Q/L) ∫ (dy cos θ/R2)
=(1/4πε0) (Q/L) ∫ (b dy / [y2 + b2 ]3/2)
(y: - L/2 → + L/2)

Ey = (1/4πε0) (Q/L) ∫ (dy y/ [y2 + b2]3/2)

The first integrale is even; then:
Ex = 2 . (1/4πε0) (Q/L) ∫ (b dy / [y2 + b2]3/2)
(y: 0 → + L/2)

The second one is odd; then
Ey = 0

I = ∫ dy / [y2 + b2]3/2 = y/ [b2(y2 + b2)1/2]
I (y: 0 → + L/2)=  (L/2)/ [b2((L/2)2 + b2)1/2]

It follows that:

E = Ex = (b Q/4πε0) / [b2((L/2)2 + b2)1/2])
With Q = λ L, we have:

E = Ex = ( λ L/4πε0) / [b((L/2)2 + b2)1/2]


 

II.3. Due to a dipole

E+ = E+ = (1/4πε0) Q / R2





R2 = a2 + x2

The projection of the vector E over the axis x gives zero.
That is : E cos θ - E cos θ = 0. It remains:
E = 2 sin θ E+ = 2 sin θ E-
That is:
E = 2 sin θ (1/4πε0) Q /  R2
We have sin θ = a/[a2 + x2]1/2
Then:
E = 2 . y /[a2 + x2]1/2 . (1/4πε0) Q / R2 
= (Q/2πε0) a / [a2 + x2]3/2
E = (Q/2πε0) a / [a2 + x2]3/2

II.4. Due to a ring

We have |dE1|  = |dE2|
dE1 is due to dQ (1) and dE1 is due to dQ (2).

We also take the opposite (2) of the dQ element in (1); then
over the axis y the components of the field cancell out; that is:
dE 1 sin θ - dE2 sin θ = 0.
It remains the sum over the x axis:
dE (total) = dE = 2 cos θ dE1 = 2 cos θ dE2
At the point P, we have: 
dE1 = (dQ/4πε0)/ [a2 + x2]1/2
Thus:
dE = (dQ/2πε0) x / [a2 + x2]3/2

E = ∫(dQ/2πε0) x / [a2 + x2]3/2
= (1/2πε0) x / [a2 + x2]3/2 ∫dQ

∫dQ = Q/2 ( Since we have considered the opposites of the elements over the ring)
It will give obviously the same result if we integrate over the whole ring without 
taking the opposites.
= (Q/4πε0) x / [a2 + x2]3/2
E = (Q/4πε0) x / [a2 + x2]3/2

II.5. Due to a charged disk

We take the last result for a ring and we integrate 
over the da element. That gives:
E = ∫ dE = (x/4πε0) ∫  dQ/ [a2 + x2]3/2
a: 0 → R
We have dQ = 2π a da σ where σ is the surface charge.
Then:
E = ∫ dE = (x/4πε0) ∫  dQ/ [a2 + x2]3/2
= (x/4πε0) ∫  [2π a da σ ]/ [a2 + x2]3/2
= (x/4 ε0) ∫  2 a da σ / [a2 + x2]3/2

∫  2 a da/ [a2 + x2]3/2 =  ∫  d(a2 / [a2 + x2]3/2 = ∫  [z + x2]- 3/2 dz = (- 2)[z + x2]- 1/2

It follows that:
E = (x/4 ε0) (- 2) σ [a2 + x2]- 1/2
a: 0 → R
E = (σ /2 ε0) [ x/|x| - x[R2 + x2]- 1/2]