Resource Lesson
Continuous Charge Distributions: Charged Rods and Rings
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Before we begin our derivations for continuous charge distributions along charged rods, let's review some relationships developed in earlier lessons.
Recall that around every charged object there is an electric field, E, with field lines that originate on positive charges and terminate on negative charges. When another charged object is brought into a field it will experience a force of attraction or repulsion according to the formula.
F = qE
When we speak of the force between two point charges, we can calculate it magnitude with Coulomb's Law
The constant, k, in Coulomb's Law has the value
where
is called the permittivity of free space and represents the "willingness" of a region to establish an electric field. This value changes with the presence of dielectrics. We can now express Coulomb's Law as
Substituting this expression for F into our equation F = qE we get an equation that we can use to calculate the magnitude of the electric field around a point charge.
In all of the following cases, the line of charge is positive.
Case I: P along the axis of a finite line of charge
Suppose you need to calculate the electric field at point P located along the axis of a finite, uniformly charged rod. Let the charge distribution per unit length along the rod be represented by λ; that is,
The total charge represented by the entire length of the rod can consequently be expressed as
Q = λL.
The charge present on a small segment of the rod, Δx
_{i,}
can be expressed as Δq
_{i}
which equals
Considering this small section to represent one of hundreds or thousands of point changes comprising the rod, its contribution to the electric field at P,
, can be represented by
The total magnitude of the electric field at P would be equal to the sum of all these smaller contributions, Δ E
_{i}
.
Taking the limit as Δx approaches 0, we get that
where x = 0 is at point P. Integrating, we have our final result of
If the charge present on the rod is positive, the electric field at P would point away from the rod. If the rod is negatively charged, the electric field at P would point towards the rod. In either case, the electric field at P exists only along the x-axis.
Simplifying our expression for E
_{P}
further we note that as b becomes much greater than L, L + b approaches b and our formula for E
_{P}
returns to the more familiar expression for a point charge
where r = b.
Case II: P along the central axis of a finite line of charge
Instead of looking for the magnitude of the electric field adjacent to one of the ends of the finite line of charge, let's now examine the electric field along a perpendicular bisector at point P as shown in the diagram below.
The charge present on a small segment of the rod, Δx
_{i,}
can be expressed as Δq
_{i}
which equals
We do not have to be concerned with the horizontal, or x-components, of the electric field at point P since symmetry reveals that they will cancel.
Consequently, all we have to do is add together all of the vertical components of the electric field at P and we will be done.
Considering this small section to represent one of hundreds or thousands of point changes comprising the rod, each
D
q
_{i}
has a vertical contribution to the electric field at point P can be represented by
Adding together all of the contributions gives us the following expressions.
Taking the limit as
D
x approaches 0, we get the expression
However, there is a problem with this integral. We cannot integrate
r
and
cos
q
with respect to
dx
. We must standardize our variables. To achieve this, we are going to use two trigonometric substitutions: one for cos
q
and the other for tan
q.
Referencing our initial charge diagram, we see that
Substituting this for 1/r in our previous equation gives us
Our expression is now slightly better since it no longer contains
r
and the constant
a
is easily removed from the integral whenever it is convenient. Our next step is to replace
dx
with
d
q
. To do this will now use the trig function tan
q
.
Isolating
dx
, gives us the expression
Now we can write our final integral and evaluate.
To finish we now need to return to our charge diagram and pay attention to assigning the limits to our integral.
Looking from left to right,
q
sweeps from
q
_{1}
to
q
_{2}
. Since we know that
q
_{1}
and
q
_{2}
have the same magnitude (remember that P is on the perpendicular bisector of our finite line of charge) we can write the relationship that
q
_{2}
= -
q
_{1 }
Since
q
_{1}
and
q
_{2 }
are equal, we can renamed our angle as just
q
without the need of any further subscripts.
But what about the length of the rod and its overall charge. How can we write an expression containing Q and L? To do this, we will use the following expression for sin
q
.
Substituting gives us
and we are finished.
Case III: P along the axis of a charged ring
Now let's assume that our finite line of charge has been curved into a circle. What would be the electric field at the center of the circle? What would be the electric field anywhere along the circle's axis?
Once again, we state with a small charge segment and add up the total affects of all segments at point P. In the following diagram, the radius of the charged ring is
a
, and the distance along the axis to P is
x
.
Once again, symmetry shows us that the y-components of the electric field will cancel, leaving only the x-components' contributions.
This derivation will be much simpler than the previous linear bisector since the values for
x
,
a
, and
r
keep the same value for all charge segments,
D
q, around the ring.
In order to simplify this integral we will use the trig function cos q.
Substituting yields
and we are finished.
Case IV: P at the center of a charged arc
Each charge segment will contribute an electric field,
D
E, at point P.
where
D
q
has been initially replaced with the expression
lD
s
and
D
s
with
a
Dq
.
Notice that the electric field at P has both x- and y-components. That is, there is no symmetry to cancel either all of the x- or all of the y-components.
We can write each of these components as
As we pass to the limit as
Dq
goes to 0 and add up all of the contributions, we get the expressions:
Thus our answer is
and we are done.
ADVANCED: P off the axis of a finite line of charge
Suppose you are now asked to calculate the electric field at point P located a distance b from the side of the uniformly charged rod.
Notice in the following diagram that we must deal with both the horizontal (E
_{x}
) and vertical (E
_{y}
) components of the electric field at P. Since the vertical components all point in the same direction -- away from the charged rod -- we will start with them.
Using vertical angles and right triangle trigonometry, we can calculate that
Taking the limit as Dx approaches 0, we get that
Unfortunately this leaves us with an expression involving three variables: x, r, and θ . Since our differentiable is dx, we need to replace r and θ with equivalent expressions involving only x. We can make this happen by noticing the following relationships:
Substituting for r and cos θ , we get
where x = 0 is at point O.
By referencing a table of integrals, we find that
allowing us to integrate and calculate our final expression for E
_{y}
.
We will now repeat our process and solve for E
_{x}
.
Using vertical angles and right triangle trigonometry, we can calculate that
Again, this leaves us with an expression involving three variables: x, r, and θ . Since our differentiable is dx, we need to replace r and θ with equivalent expressions involving only x. We can make this happen by noticing the following relationships:
Substituting for r and sin θ , we get
Since
After integrating, our final expression for E
_{x}
becomes
Some interesting consequences for this last derivation
If we place point O at
, then E
_{x}
= 0 since the left since the Δ E
_{x}
vectors cancel in corresponding pairs. This would leave us with only net E = E
_{y}
.
Suppose we view the bar from a very far distance; such that,
and
. Our expressions for E
_{x}
and E
_{y}
would reduce to
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