A thin rod of length L carries charge Q distributed uniformly over its length. (a) Show that the

step 1 have a thin rod and it's got a length L and it's got a total charge Q. What we want to do is we

A thin rod of length L carries charge Q distributed...

Key Concepts

Series Expansion and Asymptotic Analysis

Series expansion, such as a Taylor or logarithmic series, is used to approximate functions in limiting cases. In electrostatics, this approach is especially useful for deriving simplified expressions for the potential in regions far from the charge distribution, where certain terms become negligibly small, leading to expected physical behaviors like the point charge approximation.

Integration Techniques in Electrostatics

The process of integrating over a continuous charge distribution is key to obtaining expressions for the electric potential. This requires carefully accounting for distance factors between charge elements and the point of interest and often involves transforming the variables to simplify the integration limits and integrand.

Limiting Behavior and Approximation for Far Fields

Analyzing the behavior of the electric potential at large distances (r >> L) helps to verify that the derived expression reduces to a simpler, well-known form. This concept ensures that complex expressions correctly transition to expected forms in the asymptotic regime, serving as a consistency check for the solution.

Electric Potential from a Continuous Charge Distribution

This concept involves calculating the electric potential by integrating the contributions from each infinitesimal charge element of a continuous distribution, using Coulomb’s law. It forms the basis for solving problems where charges are spread over objects like rods, rings, or surfaces, and requires setting up and evaluating an appropriate integral over the charge distribution.

Uniform Charge Distribution

A uniform charge distribution means that charge is spread evenly over a given length, area, or volume. In this context, it implies that each segment of the rod contributes equally per unit length to the overall charge, thereby simplifying the integration process when calculating the electric potential and resulting field.

Transcript

00:01 Have a thin rod and it's got a length l and it's got a total charge q.

00:07 What we want to do is we want to evaluate the electric potential in a plane that is the perpendicular bisector for this rod.

00:16 So we want to know what the electric potential is at some vector are away from this rod in this perpendicular bisecting plane.

00:25 So in order to do this as usual we're going to evaluate the electric potential as being an integral over all of the charge inside our system divided by 4 pi epsilon naught curly r.

00:40 Curly r is simply the distance between the point where we're trying to evaluate the integral and the point where the charge is located.

00:50 So if i draw this system on its side we'll get a better understanding of what the geometry looks like.

00:56 So here's the rod and it's going from minus l over 2 to plus l over the 2 and let's say this is in the x direction.

01:06 If we're trying to evaluate the electric potential at this point r over here and let's consider a tiny chunk of the rod dx over here at a distance x away from the origin then this distance over here is going to be curly r.

01:25 Curly r as you can probably see from this diagram is just going to be the hypotenuse of this right angle triangle, which means that its distance is going to be square root of r squared plus x squared.

01:41 You can probably also figure out that the differential amount of charge dq is going to be equal to q over l times dx.

01:51 Q over l you might recognize is the linear charge density lambda.

01:57 Plugging all of this into our equation, we get v equals integral from minus l over 2, to plus l over 2, q over l, dx over 4 pi epsilon nought square root of r squared plus x squared.

02:16 I'm going to call 1 over 4 pi epsilon naught k, just a constant so it's easier to write.

02:23 So v is going to be equal to q over l times k integral from minus l over 2 to plus l over 2, in other words across the entire rod, dx over square root of r squared, plus x squared.

02:40 Now another thing we can do is to note that this integrand here is an even function, which means that if you put minus x or plus x, you still get the same number.

02:50 So we can simplify this integral by writing it as 2 times q times k over l, but this time making the integrand limits going from 0 to l over 2 instead.

03:06 Now all that's left to do is to evaluate this integrand here.

03:10 To do this, uh, if you it's very useful to use the substitution x equals r tan alpha.

03:20 If you do this and work through the mats, you should be able to show that the integral goes to 2q times k over l, but the limits are now from 0 to tan inverse of l over 2r, and the integrund becomes secant alpha d alpha.

03:42 If you're following along, now is the right time to pause the video and see if you can show that this substitution turns this integral into this one.

03:53 Luckily, this integral is a standard integral, which means that it's just basically 2kq over l.

04:00 And if you look at an integral table, you can see that this integral is nothing more than natural logarithm of siken alpha plus tan alpha and the limits from zero to tan inverse.

04:14 Of l over 2r.

04:18 Now, another thing we can do is to note that sqan alpha is equal to square root of 1 plus tan squared alpha.

04:26 This is a trigonometric identity that's very useful for this question, which means that when we plug in our limits, we find that the potential, as a function of r, is equal to 2k times q over l, ln, of l over 2r plus square root of 1 plus l squared, over 4r squared.

04:52 Okay, cool.

04:54 Now we're going to move on to the next part of the question, which is an important part of the question.

05:03 So i'm just going to copy the potential back again.

05:06 So here it is...

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