The electric field

The electric fieldThe electric field is the effect caused by the presence of electric charges, such as electrons, ions, or protons, in the room around him. The electric field has units of N / C or read Newton / coulomb. The electric field is generally studied in the field of physics and related fields, and indirectly also in the field of electronics that have taken advantage of the electric field in the wire conductor (cable).Origin of the electric field
Mathematical formula for the electric field can be derived through Coulomb's law, the force between two point charges: \

    \ Mathbf {F} = \ frac {q_1 q_2} {\ left | \ mathbf {r} \ right | ^ 2} \ mathbf {\ hat r}.
According to this equation, the force on a point charge is proportional to the load. The electric field is defined as a constant ratio between the charge and style [1]:

    \ Mathbf {F} = q \ mathbf {E}

    \ Mathbf {E} = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {q} {\ left | \ mathbf {r} \ right | ^ 2} \ mathbf {\ hat r}
Thus, the electric field depends on the position. A field, is a vector that depends on the other. The electric field can be considered as the gradient of the electrical potential. If some of the load is spread menghasiklan electric potential, electric potential gradient can be determined.The constant k
In common electrical formulas instead of constant k \! 1/4 \ pi \ epsilon_0 (in this paper used the last fixed), where the constant k \! The value [2]:

    \! k = \ frac {1} {4 \ pi \ epsilon_0} \ approx 8.99 \ times 10 ^ 9 N m2 C-2
the so-called equality of force constant power [3].Calculate the electric fieldElectric Field.png
To calculate the electric field at a point \! \ Vec {r} due to a point charge \! q which is located in \! \ Vec {r} _q used the formula [4]

    \ Vec {E} (\ vec {r} - \ vec {r} _q) \ equiv \ vec {E} (\ vec {r}; \ vec {r} _q) \ equiv \ vec {E} (\ vec {r}) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {q} {\ left | \ vec {r} - \ vec {r} _q \ right | ^ 3} \ left (\ vec {r} - \ vec {r} _q \ right)
Simplification of the lack of proper
Generally, to simplify selected centers coincide with the coordinates of the point charge \! q which is located in \! \ Vec {r} _q to obtain such formulas have been written at the beginning of this article, or if rewritten in vector notation:

    \ Vec {E} (\ vec {r}) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {q} {\ left | \ vec {r} \ right | ^ 3} \ vec {r }
with the unit vector \! \ Hat {r}
\ Hat {r} = \ frac {\ vec {r}} {\ left | \ vec {r} \ right |} = \ frac {\ vec {r}} {r}.
It is recommended to use a formula involving \! \ Vec {r} _q and \! \ Vec {r} for more general and can be applied to the case of more than one charge and also the charge distribution, both discrete and continuous distributions. Simplification is also sometimes make understanding calculate the electric field in a little bit blurred. In addition it is also because of this simplification is only one special case in the calculation of the electric field (case by a point charge at which point the charge is placed at the origin).Sign of electric chargeElectric Field Lines.svg
Electric charge can be negative, zero (there is no charge or a number of units equal positive and negative charge) and negative. The value of this charge will affect the calculation of the electric field in terms of the sign, which is positive or negative (or zero). If at any point around a (or several) calculated charge and electric field vectors described, it would seem the lines are interconnected, which is called the electric field lines. Sign of the charge to determine whether the electric field lines are caused comes from him or towards him. It has been determined (by the force experienced by a positive test charge), that

    positive charge (+) will cause the electric field lines directed from him toward the exit,
    negative charge (-) will cause the electric field lines directed toward the entrance to her.
    zero load () does not cause the electric field lines.
Electrical potential gradient
The electric field can also be calculated when an electric potential \! U is known, by calculating the slope [5]:

    \ Vec {E} = - \ vec {\ nabla} U
with

    \ Vec {\ nabla} = \ hat {i} \ frac {\ partial} {\ partial x} + \ hat {j} \ frac {\ partial} {\ partial y} + \ hat {k} \ frac {\ partial} {\ partial z}
for the Cartesian coordinate system.Electric field energy
The electric field stores energy. Meeting energy an electric field is given by [6]

    u = \ frac {1} {2} \ epsilon | E | ^ 2
with

    \ Epsilon \! permittivities are medium in which the electric field there, the vacuum \ epsilon = \ epsilon_0 \!.
    E \! is the electric field vector.
Total energy stored in the electric field within a volume V \! is

    \ Int_ {V} \ frac {1} {2} \ epsilon | E | ^ 2 \, d \ tau
with

    d \ tau \! is the differential element of volume.
Distribution of electric charge
The electric field need not only caused by the electric charge, but can also be caused by more than one electric charge, even by the electrical charge distribution both discrete and continuous. Examples such as electric charge distribution:

    charge collection points
    long straight wire of finite and infinite
    circular wire
    finite width plate or infinite
    thin discs and rings
    other forms
Charge collection points
To charge points and totals are not spread too much, the electric field at a point (and not at one point charge) can be calculated by summing the electric field vector at that point due to each charge. In this case it is better written

    \ Vec {E} _i (\ vec {r}) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {q_i} {\ left | \ vec {r} - \ vec {r} _i \ right | ^ 3} \ left (\ vec {r} - \ vec {r} _i \ right)
is read, the electric field at point \ vec {r} due to the charge \! q_i located in \ vec {r} _i. Thus the electric field at point \ vec {r} caused the entire load scattered written asElectric field 4 point charges 1.png

    \ Vec {E} (\ vec {r}) = \ sum_ {i = 1} ^ {N} \ vec {E} _i (\ vec {r})
where \! N is the number of charge points. As an illustration, for example, to extrapolate the magnitude of the electric field at the point \! P which is the second intersection of the diagonal of a square-sided \! R, in which there is a charge made by the four points lie on the vertex-vertex of the square. For this case suppose that q_1 = q_3 = + Q \! and q_2 = q_4 =-Q \! and grab the center point coordinates in \! P (0.0) for convenience. For two-dimensional cases such as this, can be written also

    \ Vec {E} _i (\ vec {r}) = \ vec {E} _i (x, y)
which will provide

    \ Vec {E} _1 (0.0) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {Q} {\ left (\ frac {R} {4} ^ 2 + \ frac {R} {4} ^ 2 \ right)} \ \ frac12 \ sqrt2 (\ hat i - \ hat j)

    \ Vec {E} _2 (0,0) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {Q} {\ left (\ frac {R} {4} ^ 2 + \ frac {R} {4} ^ 2 \ right)} \ \ frac12 \ sqrt2 (\ hat i + \ hat j)

    \ Vec {E} _3 (0,0) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {Q} {\ left (\ frac {R} {4} ^ 2 + \ frac {R} {4} ^ 2 \ right)} \ \ frac12 \ sqrt2 (- \ hat i + \ hat j)

    \ Vec {E} _4 (0,0) = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {Q} {\ left (\ frac {R} {4} ^ 2 + \ frac {R} {4} ^ 2 \ right)} \ \ frac12 \ sqrt2 (- \ hat i - \ hat j)
so that

    \ Vec {E} (0.0) = \ sum_ {i = 1} ^ {4} \ vec {E} _i (0.0)

    \ Vec {E} (0.0) = \ vec {E} _1 (0,0) + \ vec {E} _2 (0,0) + \ vec {E} _3 (0,0) + \ vec { E} _4 (0.0)

    \ Vec {E} (0.0) = \ vec {0}
that generates the electric field at that point is zero.Long straight wireLine charge.png
Long, straight wire is one form of the charge distribution is of interest because if taken infinite length, load calculations on a range of wire and located in the middle of its length, it becomes very easy.
For a wire that stretches straight on axis x \!, At a distance z \! on it, with wires extending from-a \! to b \! from the point of projection P \! the wire, the electric field at that point can be calculated magnitude, namely:

    E_z = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {\ lambda} {z} \ \ left [\ frac {b} {\ sqrt {z ^ 2 + b ^ 2}} + \ frac { a} {\ sqrt {z ^ 2 + a ^ 2}} \ right]
As mentioned above, if-a \ rightarrow - \ infty and b \ rightarrow \ infty then using L'Hospital theorem obtained

    E_z = \ frac {1} {4 \ pi \ epsilon_0} \ \ frac {2 \ lambda} {z} = \ frac {\ lambda} {2 \ pi \ epsilon_0z}
Or if the wire is placed parallel to the z-axis and the xy plane perpendicular wire penetrated, then the electric field at a point within \! R from the wire, is the electric field can be written

    \ Vec {E} (r) = \ frac {\ lambda} {2 \ pi \ epsilon_0r} \ hat {\ rho}
with \ hat {\ rho} is the radial unit vector in cylindrical coordinates:

    \ Hat {\ rho} = \ hat {i} \ cos \ phi + \ hat {j} \ sin \ phi
where \ phi \! is the angle formed by the positive x-axis.