Electric Field and Potential by Direct Integration
A
point charge produces an electric field
given by the formula 
and
a electric potential given by the formula, 
,
where Q is
the charge and 
is
the vector distance from the charge to the field point P where we want to
calculate the field or potential (see Figure 1). We find the electric
field or
potential due to a charged wire by considering it to be made up of many
infinitesimally small pieces of length dS each
containing charge dQ
(see Figure 2). Note that will
be different for each different piece of the wire. We will
need to sum up each
infinitesimal contribution via an integral.
There are a number of steps to properly constructing the integral for a straight wire.
· First note that your answer can only have the symbols given in the question.
· Choose a coordinate system and origin. If there is a lot of symmetry, put the origin at the symmetry point.
· If the wire lies on the x axis in your coordinate system, then choose an arbitrary piece of the wire and say it is distance x from the origin – x is your variable of integration. The size of the piece will then be dx. Note that x should be an arbitrary positive point – not the ends or middle.
·The limits of integration are the values of x for the ends of the wire.
The piece dx
will have charge 
if
you are given Q and L or 
if
you are given the charge density, λ.
·
The vector
distance should
be expressed in i j k notation.
In general each component
of will
depend the variable x.
·
The magnitude
of the vector ,
R, is found using the Pythagorean identity 
.
·
The electric
field is thus given by 
.
· If symmetry indicates that one of the components is zero, you should indicate which it is.
·
The electric
potential is given by 
Example
Find both the electric field and electric potential due to a straight wire of length L carrying uniform charge Q at a point P a distance b to the right and 2b above the right end of the wire as shown in the diagram below.
Solution
First set up the axis and identify limits of integration.
Second,
choose an arbitrary piece of the
wire of length dx, label its location as x,
and its charge as or
depending
on the given information.
Third,
identify in
the drawing and express it in i j k
notation.
Your electric field is thus
and the potential is
.
Curves
If the wire is a circular arc with radius or curvature r then the natural variable of integration is the angle θ measured from the centre of the radius or curvature. The limits of integration will be start and end angles of the wire.
There are a number of steps to properly constructing the integral for a curved wire.
· First note that your answer can only have the symbols given in the question.
· Choose a coordinate system and put the origin at the centre of the radius of curvature.
· Choose an arbitrary piece of the wire at an angle θ from the positive x axis – θ is your variable of integration. The size of the piece will then be dS = rdθ where r is the radius of curvature. Note that θ should be an arbitrary positive angle – not the ends or middle.
· The limits of integration are the values of θ for the ends of the wire.
·
The piece dS
will have charge 
if
you are given Q and L or 
if
you are given the charge density, λ.
·
The vector
distance should
be expressed in i j k notation.
In general each component
of will
depend the variable θ usually in the form of cosθ
or sinθ.
·
The magnitude
of the vector ,
R, is found using the Pythagorean identity .
·
The electric
field is thus given by 
.
· If symmetry indicates that one of the components is zero, you should indicate which it is.
·
The electric
potential is given by 
Example
Find both the electric field and
electric potential due to a
curved semicircle of wire of radius r carrying
uniform charge density λ at a point P
which a distance 2b to the right and b
above the right end of the
wire as shown in the diagram below.
θ
Solution
First set up the axis with the origin
at the centre of curvature.
or
depending
on the given information. Identify the limits of integration from the
start and end angles of the wire.
in
the drawing and express it in i j k
notation using cosθ
and sinθ.
and the potential is
.