28: PSS 25.2 The Electric Potential of a Continuous Distribution of Charge

INTRO:
To practice Problem-Solving Strategy 25.2 for continuous charge distribution problems.
A straight rod of length L has a positive charge Q distributed along its length. Find the electric potential due to the rod at a point located a distance d from one end of the rod along the line extending from the rod. (Figure 1)

Figure 1

PROBLEM-SOLVING STRATEGY 25.2 The electric potential of a continuous distribution of charge
MODEL: Model the charges as a simple shape, such as a line or a disk. Assume the charge is uniformly distributed.
VISUALIZE: For the pictorial representation:
Draw a picture and establish a coordinate system.
Identify the point P at which you want to calculate the electric potential.
Divide the total charge Q into small pieces of charge ΔQ using shapes for which you already know how to determine V. This division is often, but not always, into point charges.
Identify distances that need to be calculated.
SOLVE: The mathematical representation is V=ΣVi.
Use superposition to form an algebraic expression for the potential at P.
Let the (x,y,z) coordinates remain as variables.
Replace the small charge ΔQ with an equivalent expression involving a charge density and a coordinate, such as dx, that describes the shape of charge ΔQ. This is the critical step in making the transition from a sum to an integral because you need a coordinate to serve as an integration variable.
Express all distances in terms of the coordinates.
Let the sum become an integral. The integration will be over the coordinate variable that is related to ΔQ. The integration limits for this variable will depend on the coordinate system you have chosen. Carry out the integration, and simplify the result.
ASSESS: Check that your result is consistent with any limits for which you know what the potential should be.

Model
No information is provided on the rod's cross section. For simplicity, assume that its diameter is much smaller than the rod's length. You can, then, model the rod as a line of charge.

Visualize
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PART A:
To most efficiently solve this problem, you should divide the rod into pieces of charge that consist of...

- thin lines of charge of length L  and a very small cross section
- thin 'slices' of the rod cut parallel to the axis of the rod
- thin 'slices' of the rod cut perpendicular to the axis of the rod

SOLUTION:
First, model the rod as a line of charge. Then, divide the line into many small segments, each of length Δx. Each segment can, then, be modeled as a point charge. The potential due to such a point charge can be determined in a relatively straightforward manner.

∴ option #3, thin 'slices' of the rod cut perpendicular to the axis of the rod
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PART B:
Choose a coordinate system where +x is to the right and +y is upward. Place the origin of your coordinate system at the left end of the rod, and choose point P to be located beyond the right end of the rod, as shown in the picture to the left.
What is the distance ri between point P and a piece of charge located at position xi?
Express your answer in terms of some or all of the quantities xi, L, Q, and d.

SOLUTION:
A pictorial representation for this problem might look like:

r_i = (L+d)-x_i
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PART C:
Find the electric potential V_P at point P.
Express your answer in terms of d, L, Q, and ϵ₀.

SOLUTION:
The mathematical expression for the potential is V=ΣVi. 
Because of the nature of the problem, which deals with a continuous distribution of charge rather than a point-like distribution, the sum must be treated as an integral. 

To set up the summation and find what to integrate, follow the steps listed in the strategy ^^

After dividing the total charge into small segments, and modeling each segment as a point charge, as in Part A, construct a mathematical expression for Vi, the electric potential due to segment i. 
Treat xi, the x coordinate of segment i, as a variable (just call it x), and express the charge on that segment in terms of said x. 

Finally, let ΣVi become an integral and use calculus to evaluate the integral...

FIND... the electric potential Vi at point P due to the charge segment located at the variable position 
x

The formula for the electric potential of a point charge involves the distance from the charge to the point where the potential is to be computed.
Recall that, in Part B, you obtained an expression relating the distance between point P and charge segment i to the x coordinate of the charge segment.......

The formula for electric potential is... 
U_elec = k q₁q₂/r
and we all know what k is so that will be temporarily ignored...

The immediate distance between the two points is r_i, as determined in part B... 

U = k q₁q₂/r ...  then just plug in for r as r_{i ...}
V_i = k ⋅ΔQ/(L+d-x_i)

Next you need to understand the expression for the charge of a small segment
So calculus is kind of annoying at first, but these topics are actually really important to understand. especially later on (trust me)..

Because you will integrate with respect to coordinate x, it is important to express the charge on a segment, ΔQ, in terms of its length Δx.

So you gotta find a mathematical expression for ΔQ, while assuming the charge is uniformly distributed along the rod (since that's given). 

Recall that the total charge on the rod is Q and the rod's length is L...
that means that the change in charge over the change in x is simply the charge divided by the length, or...
ΔQ/Δx = Q/L
by rearranging this to solve for ΔQ...

ΔQ = Δx⋅Q/L

plug that into your initial V_i equation for ΔQ, 
V_i = k ⋅Δx⋅Q/L ⋅/(L+d-x_i)
you now have an integrable function..
V_i = ∫k⋅Q/L⋅(L+d-x_i) Δx = ∫k⋅Q/L⋅(L+d-x) dx
integrate from 0 to L, because the d is already attributed... 

k is constant, Q is constant, L is constant, so you can just pull that out...
→k⋅Q/L ⋅∫1/(L+d-x) dx
use substitution...

let u = L+d-x, ∴du = -dx

→k⋅Q/L⋅∫1/(u) -du
→k⋅Q/L⋅-ln(u) (from 0 to L)usub
→k⋅Q/L⋅-ln(L+d-x) (from 0 to L)
→k⋅Q/L⋅[-ln(L+d-L) - -ln(L+d-0)]
→k⋅Q/L⋅[-ln(d) + ln(L+d)]
using law of logs, subtracting them is the same as dividing them...
→k⋅Q/L⋅ln[(L+d)/d] = k⋅Q/L⋅ln[L/d +d/d]

⇒V_p = k⋅Q/L⋅ln[1+ L/d] = Q/4Lπε₀⋅ln[1+ L/d]

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PART D:
Imagine that distance d is much greater than the length of the rod. Intuitively, the potential should be approximately the same as the potential at a distance d from which of the following charge distributions?

- an infinitely long wire with total charge Q
- an infinitely long wire with total charge Qd/L
- a point charge of magnitude Q
- an electric dipole with moment QL

SOLUTION:
From far away, a short rod looks very much like a simple point charge. Not surprisingly, the mathematical expression you obtained for the potential does reduce to that of a point charge if L/d≪1:

In other words, since L is constant, the larger d is, the smaller L/d gets. When L/d is really small, the logarithm is basically just the logarithm of 1 plus that tiny little amount. The ln(1) = 0, and oddly enough, at really close proximity to 1, those values follow the rule :

if x << 1 & ln(1+x), then ln(1+x) ≈ x​

try this with x = 0.001, for example... ln(0.001) = 0.0009995 ≈ 0.001

therefore, Q/4Lπε₀⋅ln[1+ L/d] where L/d << 1 ≈ Q/4Lπε₀⋅L/d 
⇒ or, ≈ Q/4dπε₀

Therefore, the third option is correct. or, 

a point charge of magnitude Q​