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## Homework Statement

[/B]

Thin rod of length [itex]l[/itex] is placed with one of its ends at the center [itex]O[/itex] of the (thin circular )ring of radius R as shown (Figure 1), perpendicular to the plane of the ring. The rod is charged with total charge Q that is distrubted along the rod's length with linear charge density

__linearly__proportional to the distance from the bottom end of the rod. The ring is uniformly charged with total charge [itex]q_0[/itex].

**Find the electrostatic interaction energy between the rod and the ring.**

(Figure 1)

## Homework Equations

(Definition of Interaction Energy)[/B]

$$W_{\text{interaction}}=\varepsilon_0\int_{\text{all space}}\vec{E}_{\text{rod}}\cdot\vec{E}_{\text{ring}}d\tau$$

**(Interaction Energy of two point charges q**

_{a}and q_{b})$$W_{\text{interaction}} = \frac{q_a q_b}{4\pi\varepsilon_0r^2}$$

I'd imagine that finding expressions for the field of these two objects would be very cumbersome because there is little to no symmetry to exploit. We could find expressions for these fields easily if we limited ourselves to the surface of the ring or the axis of the rod, but the definition of interaction energy requires that we have expressions for field that apply to all space.

Instead I'd like to treat the system as a collection of point charges and appeal to the second equation above.

## The Attempt at a Solution

[/B]

$$dq_{rod} = \frac{2Qz}{l^2} dz$$

where dz is a differential element of length for the rod, notice that if we integrate dq

_{rod}from [itex]0[/itex] to [itex]l[/itex] we get the total charge of Q

$$dq_{ring} = \frac{q_0 dc}{2 \pi R}$$

where dc is the differential element of length(/circumference) for the ring

The interaction between a differential element of length for the rod and a differential element of length for the thin circular wire is

$$d\zeta = \frac{dq_{rod} dq_{ring}}{4\pi\varepsilon_0 (R^2 + z^2)} $$

plugging in the expression for dq

_{rod}and dq

_{ring}we get

$$ d\zeta = \frac{q_0 Q z dc dz}{4\pi^2\varepsilon_0 R l^2 (R^2 + z^2)}$$

If we (double) integrate this expression over ##0 \leq c \leq 2piR## and ##0 \leq z \leq l## then we should get the total interaction energy between the ring and the rod

$$\int_{z=0}^l \int_{c=0}^{2 \pi R} \frac{q_0 Q z dc dz}{4\pi^2\varepsilon_0 R l^2 (R^2 + z^2)}, dc\, dz$$

$$ = \frac{q_0 Q}{2\pi\varepsilon_0l^2} \int_{z=0}^l \frac{z}{R^2 + z^2} \, dz$$

$$ = \frac{q_0 Q}{4\pi\varepsilon_0l^2} \left (\ln{|R^2 +z^2|}) \right|_0^l$$

for a

**final answer**of

$$\zeta = \frac{q_0 Q}{4\pi\varepsilon_0 l^2} \ln{\left|\frac{R^2 + l^2}{R^2}\right|} $$

**(TOTAL INTERACTION ENERGY BETWEEN ROD AND RING)**

**4. Additional Comments**

My friend contends that I am wrong because I applied the product rule incorrectly to dq

_{ring}dq

_{rod}. He had a similar problem for homework about 2 years ago and he did it using the expression for fields which to me is cumbersome at best and impossible at worst. In fact he posted the same question in this section around that time https://www.physicsforums.com/threa...-energy-between-a-charge-rod-and-ring.770828/.

This is my first time posting as well as my first time using Latex so I apologize if I have unknowingly broken any rules and for formatting errors. I know my username is immature; I made it a few months ago before upper level physics forced me to grow up.

I look forward to participating in this community and I would like to say thank you in advance for any help and comments.