Lecture 4 - ENERGY AND POTENTIAL

ENERGY AND POTENTIAL
Energy expended in moving a point charge in an electric field
Suppose we wish to move a charge a distance in an

electric field . The force on due to the electric field is
Where the subscript reminds us that this force is due

to the field. The component of the force in the
direction which we must overcome is
, where is the unit vector in the direction of .
The force we must apply is equal and opposite to the
force due to the field.
And our expenditure of energy is the product of the
force and distance and the differential work
performed by external source moving .
The work done by an external force in moving a point

charge from some point to a point is
Where the integral from is a line integral over a

specified path from point to a point .
POTENTIAL DIFFERENCE (POTENTIAL) ABOUT A POINT CHARGE
The potential difference between points , is

defined as the work in joules that we must
expend in moving a unit positive point charge
from point in an electric field .
In equation form;
Equation (1) is a line integral over a specified

path from .
In evaluating line integrals, the differential
distance is always positive, even though the
path is directed in a decreasing coordinate
value.
The expression for will take on one of the
following forms, depending on the coordinate
system selected;
Commonly, the potential difference is referred to as the
Potential of point . This usage must be backed up with a
statement of the reference point since the potential of
point depends on the reference selected.
When the reference is taken to be at infinity, the

potential is referred to the Absolute Potential. It should
be noted that the absolute potential is a potential
difference with reference point at infinity.
Let us find the potential difference about a positive point

charge at the origin as shown
Let us first find by integrating (1) over a general
path from point to point to obtain
From (2), it can be seen that from a static point
charge depends on the end point coordinates
and is thus independent of the path taken from
point to point .
An field whose integral is independent of path
will yield
Equation (3) indicates that zero amount of work
is done by us if we carry a about any closed
path. This conservation of energy property of
the static field prompts us to call this field a
Conservative Field.
Now
If becomes infinite we get

is called the absolute potential or just the
potential of point , referenced to infinity.
Equation (4) indicates that any point on a
surface of constant radius, , will have the same
potential. Surfaces of the same potential will be
called equipotential surfaces.
From (4)
From (5), it can be seen that we can evaluate the
potential difference by taking the difference of
potentials, found at points , if they are referenced
to the same point; Infinity in this case.
Example
Find by integrating over path of the figure below
in an electric field when the points are at
respectively.
Solution
In rectangular coordinates
Over the path , we find that and . Over the

path , we find that and .
Example 2
Repeat the example above over the path , a
straight line from .
Solution
Over the path , we find that

Alternatively;
From the two examples above it can be seen

that the field is a conservative field and it
requires of energy to carry a of charge from
point .
TRY
Evaluate the potential in an electrostatic field .
When the path of integration is
A series of straight lines from .
A straight path from .
[Ans: ]
POTENTIAL DIFFERENCE (POTENTIAL) ABOUT A CHARGE SYSTEM
For the system of finite point charges shown

below, the absolute potential at point , becomes
Where are the electric fields due to the point charges acting
alone. Equation (1) becomes
Where are the distances from to the point respectively.

Equation (2) can be rewritten as
Where are the absolute potentials at point due to
acting alone.
If we let in (2) and increase the number of point

charges to infinity, we have
Where is the distance from to point , and the

integral is carried out over the charge in the system.
From (4)
Equation (4) takes on the following forms for
line, surface and volume charge distributions.
Equations (1) and (7) can be written using
primed variables for charge locations and
unprimed variables for field locations, thus
And
It should be noted that in (8) is a differential
vector distance along a path from in space while
the in (5) is the differential length along a line of
charge.
We now have two distinct methods (8) and (9)
to find the absolute potential.
Equation (8) requires knowledge of the while (9)
requires the knowledge of the charge
distribution.
Example
The electric field along the axis of a ring of
uniform and whose radius is is given by
Find the absolute potential along the z-axis.
Solution
axis
𝑑𝑥=𝑑𝑦 =0 𝑎𝑛𝑑 ⃗
𝑑𝑙= ^
𝑧 𝑑𝑧
Example 2
For the previous example, find using equation (5).
Solution
along the line of charge and is the distance from

the location of the point charge to the point on
the z-axis where the absolute potential is to be
found.
GRADIENT OF A POTENTIAL DIFFERENCE
Now
Now,
From (1) and (2)

The expanded expression for in rectangular,
cylindrical and spherical coordinate systems are
given below
Example
By using , find the field about a point charge at
the origin when
Solution
Now
Since is not a function of .

Lecture 4 - ENERGY AND POTENTIAL

Uploaded by

Copyright:

Available Formats

Lecture 4 - ENERGY AND POTENTIAL

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Lecture 4 - ENERGY AND POTENTIAL

Uploaded by

Copyright:

Available Formats

ENERGY AND POTENTIAL

Energy expended in moving a point charge in an electric field

Suppose we wish to move a charge a distance in an

Where the subscript reminds us that this force is due

The work done by an external force in moving a point

Where the integral from is a line integral over a

The potential difference between points , is

Equation (1) is a line integral over a specified

When the reference is taken to be at infinity, the

Let us find the potential difference about a positive point

If becomes infinite we get

Over the path , we find that and . Over the

Over the path , we find that

From the two examples above it can be seen

For the system of finite point charges shown

Where are the distances from to the point respectively.

If we let in (2) and increase the number of point

Where is the distance from to point , and the

along the line of charge and is the distance from

From (1) and (2)

You might also like