Electrostatics
Electromagnetics
Electricity & Magnetism
© The scientific sentence. 2010
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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
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
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x
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V(x) = - ∫ | |
E (x') dx'
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∞
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=
=
a2kq/x3
V(x) = a2kq/x3
That we can find also by the following equation:
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