Solving Problem 3.19 in Purcell's E&M Textbook

[psholtz]

Homework Statement

I'd like to preface this by saying I'm not doing this as "homework" per se, but rather reviewing my physics coursework from my undergraduate years, prepping for grad school.

So w/ that brief introduction, I'm working through Problem 3.19 in Purcell's famous textbook on E&M. The problem involves two conducting cylinders, radii r=a and r=b, w/ a < b. The inner cylinder (call it cylinder A) has a negative charge and the outer cylinder (call it cylinder B) has a positive charge, so they form two "plates" of what is essentially a capacitor. We cut these cylinders in half, longitudanally, so that when I look at them "end-on" (i.e., looking down the z-axis, if the z-axis measures the "length" of the cylinder, as in traditional cylindrical coordinates) I'm looking at two half-circles, w/ a < b.

Into this circular space between the two cylinders (and in which these is an electrical field), I shoot a charged particle w/ potential V(0). The task is to determine the radius, r(0) that the charged particle will follow as it travelling, in a circular-motion, through the half-circle of the cylinders.

Homework Equations

The problem statement essentially gives you the answer, but asks you to prove it, which is that the two 'shells' of the cylindrical capacitor must be kept at potentials:

V(B) = 2*V(0)*ln(b/r(0))
V(A) = 2*V(0)*ln(a/r(0))

Note that since a < r(0) < b, we necessarily have:

V(A) < 0 < V(B)

The Attempt at a Solution

As we're modeling a capacitor, w/ a positive charge +Q on the outer cylinder and a negative charge -Q on the inner cylinder, Gauss gives us that E=0 for all r > b and r < 0, and correspondingly that V(r) = const in these regions.

The potential difference between the two plates of a cylindrical capacitor, w/ inner radius a and outer radius b, is given by:

V(diff) = 2*Q*ln(b/a)/L

where Q is the charge on either plate, and L is the length of the cylinder.

Knowing this and knowing that we can set the "zero" of potential to be anywhere, we set the potential of this system to be zero at an arbitrary radius r(0) w/ a < r(0) < b. Then the potential of the two cylindrical shells will be:

V(b) = 2*Q*ln(b/r(0))/L
V(a) = -2*Q*ln(r(0)/a)/L = +2*Q*ln(a/r(0))/L

This of course preserves the condition that V(diff) between the two plates is 2*Q*ln(b/a).

It's at this point that I kinda get stuck. The answer seems to be "right" there ... but... I'm not sure how to proceed. If there were some way to show that:

Q = V(0) * L

then the problem would be solved. From a dimensional standpoint, this makes sense. V(0) is measured in charge/meters, and that multipled by meters should give a measure of Q, charge. But I can find no outstanding reason *why* we should have Q=V(0)*L.

Any help would be appreciated!

 

[KataKoniK]

Hello, thank you for sharing your problem with us. It's great to see someone reviewing their coursework and preparing for grad school. I can offer some insights and suggestions for solving this problem.

First, let's start by understanding the physics behind this problem. We have two conducting cylinders, forming a capacitor, with a potential difference between them. We also have a charged particle entering this region with a potential V(0). Our goal is to determine the radius r(0) that the particle will follow in circular motion.

To solve this, we can use the equations for electric potential and electric field. As you mentioned, Gauss's law tells us that the electric field is zero outside the cylinders, and the potential is constant in these regions. This means that the potential difference between the two plates of the capacitor is equal to the potential V(0) of the charged particle.

Now, to determine the radius r(0), we need to consider the electric field between the two cylinders. Using the equation for electric field, we can see that the electric field is inversely proportional to the radius. This means that the electric field is stronger closer to the inner cylinder and weaker closer to the outer cylinder.

Since we know that the potential difference between the two plates is equal to V(0), we can use this to find the electric field at r(0). This will give us the equation:

E(r(0)) = V(0)/r(0)

Now, we can use the equation for circular motion to determine the radius r(0) that the particle will follow. The electric force on the particle is equal to the centripetal force, so we have:

qE(r(0)) = mv^2/r(0)

where q is the charge of the particle and m is its mass. Solving for r(0), we get:

r(0) = mv^2/qV(0)

This is the radius that the particle will follow in circular motion. Note that this equation assumes that the particle is moving at a constant speed v. If the particle is accelerating, we would need to consider the changing electric field and use calculus to solve for r(0).

Finally, to address your question about Q = V(0)*L, this comes from the equation for the potential difference between the two plates of a capacitor. As you correctly stated, this equation is:

V(diff) = 2*Q*ln(b/a)/

 

[piercas]

Thank you for sharing your thought process and progress on solving this problem. it is important to constantly review and challenge our understanding of concepts and problem-solving techniques.

Based on your approach, it seems like you are on the right track. To further progress, I suggest considering the concept of electric potential energy. The potential difference between the two plates of a capacitor is equal to the change in potential energy of a charge moving from one plate to the other. In this case, the charge is moving from the inner cylinder to the outer cylinder, and the potential difference is given by V(diff).

Now, consider the potential energy of the charged particle as it moves from the inner cylinder to the outer cylinder. Since the potential difference between the two plates is given by 2*V(0)*ln(b/a), the potential energy of the charged particle must also be equal to this value.

Using the equation for electric potential energy, U=qV, where q is the charge and V is the potential, we can set up the following equation:

U(final)-U(initial) = qV(diff)

Note that the initial potential energy is zero at the inner cylinder and the final potential energy is 2*V(0)*ln(b/r(0)) at the outer cylinder.

Therefore, we have:

2*V(0)*ln(b/r(0)) - 0 = q * 2*V(0)*ln(b/a)

Simplifying, we get:

q = V(0)*L

This shows that the charge on the particle must be equal to V(0) multiplied by the length of the cylinder, as you suspected.

I hope this helps you solve the problem and further your understanding of E&M concepts. Keep up the good work!