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© The scientific sentence. 2010

Electrostatics: Electric field



1. Electric field


A vector field is vectorial quantity at each point in space.

A mass (earth) produces a gravitational field around it, defined as the ratio F/m, where F is the force (weight) exerted on an object test of mass m.

This ratio if equal to g called gravitational field of magnitude 9.8 N/kg.

Similarly, a charge (q) placed somewhere in space produces an electric field around it in space of magnitude E = F/qo, where F is the force exerted by the charge q on the charge test qo.

Applying Coulomb's law gives

E = F/qo

  k q   ^
E =   
  r
   r2  


The electric field does not depends on the charge test qo.



More generally, the net electric field E in space where are present n charges q1, q2, ..., qi, ..., qn is given by:

   n k qi     ^
E  =  Σ  
    ri
    i   ri2  




The electric field due the set of charge particles depends on the value of the charge on each paricle and how the particles are presents in space, that is the charge distribution, and the position where to measure the electric field. The charge of the test particle must be small to not alter the charge distribution or point charges.

The direction of the vector field E is the direction of the force on a positive charged test paticle.

The dimension of the electric field in the SI unit is Newtons by Coulomb (N/C).



2. Example

Lat's onsider this quadrupole of four points charges q1 = q2 = q3 = q4 = q located respectively at (- a, 0), (+ a, 0), (0, + a) and (0, - a).

We want to determine the electric field at the distance x on the x-axis from the origine o.

The electric field, as the force, obeys the principe of superposition, that is the net electric field E is the vectorial sum of the related components.

Vectorially, E = E1 + E2 + E3 + E4.

The magnitude of the related field are:

E1 = k q1/(x + a)2 = k q/(x + a)2
E2 = k q2/(x - a)2 = k q/(x - a)2
E3 = k q3/(x2 + a2) = k q/(x2 + a2)
E4 = k q4/(x 2 + a2) = k q/(x 2 + a2)

E1 and E2 are along the x-axis. They are positive.

E3 and E4 have two components each:
E3(E3x, E3y), and E4 (E4x, E4y).

E3y and E4y cancel.
It remains the two components E3x and E4x.

E3 = E4 = k q/(x2 + a2)

E3x = E4x = - E3 cos θ

cos θ = x/[x2 + a2]1/2

E3x + E4x = - 2 E3 cos θ = - 2 E3 x/[x2 + a2]1/2
= - 2 k q x/(x2 + a2)/[x2 + a2]1/2 =
- 2 k q x/(x2 + a2)3/2

E3x and E4x are along the x-axis. They are negative.

Therefore:

E = E1 + E2 - E3x - E4x =
k q/(x + a)2 + k q/(x - a)2 - 2 k q x/(x2 + a2)3/2 =
k q[1/(x + a)2 + 1/(x - a)2 = - 2 x/(x2 + a2)3/2]
k q[1/(x + a)2 + 1/(x - a)2 - 2 x/(x2 + a2)3/2]

We have :

1/(x + a)2 = 1/x2(1 + a/x)2
1/(x - a)2 = 1/x2(1 - a/x)2
x/(x2 + a2)3/2] = x/x3(1 + a2/x2)3/2] =
1/x2(1 + a2/x2)3/2]

Therefore

E = (kq/x2)[ 1/((1 + a/x)2 + 1/(1 - a/x)2 - 2/(1 + a2/x2)3/2]

The binomial expansion second approximation
(1 + ε)n ≈ 1 + nε , with ε << 1, gives:

E = (kq/x2)[ 1 - 2(a/x) + 1 + 2(a/x) - 2(1 - (3/2) a2/x2)] =
(kq/x2)[ 2 - 2 + 3 a2/x2] = (kq/x2)[ 3 a2/x2] =
3 a2kq/x4)

E(x) = 3 a2kq/x4



2. the related potential

  x  
V(x) = - ∫   E (x') dx'
   
=
  x  
- ∫   3 a2kq/x'4 dx'
   

= a2kq/x3

V(x) = a2kq/x3

That we can find also by the following equation:

   4 k qi     ^
E  =  Σ  
    ri
    i   ri  











 


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