Force due to two thin charged rods acting on each other

[Vriska]

Homework Statement

I have 2 thin rods of length L in the axial plane(x axis), they're of the same uniform linear charge distribution and are separated by a distance a .

Homework Equations

$$E = \frac{kq} {R^2}$$
$$F = qE$$

The Attempt at a Solution

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Let 0 be one at one end of the two rod systems such that the first rod terminate at L and the other rod begins at L+ a.

Then field E due to first rod at some point a =
$$\int \frac{\lambda k dx}{(L + a - x)^2} $$ evaluated from x = 0 to x = L. This is $$\frac{L}{a(L+a)}$$. Now I find the force on the second charge $$ dF = E'(x) dq$$ we note here that E'(x) = E(a +x) integrating this from 0 to L. gives you the wrong answer : (

Ps. Where can I find more of these infinitismal problems for practice?

 

[haruspex]

Then field E due to first rod at some point a =
$$\int \frac{\lambda k dx}{(L + a - x)^2} $$

a is not some arbitrary point. Your integrand appears to be the field at the start of the second rod due to the first rod, i.e. its field at L+a.
You need the field due to the first rod at an arbitrary point within the second rod.

 

[Vriska]

a is not some arbitrary point. Your integrand appears to be the field at the start of the second rod due to the first rod, i.e. its field at L+a.
You need the field due to the first rod at an arbitrary point within the second rod.

It's the field at the start of the second rod but I'm increasing the distance a by some dx and summing it till 2L+a. Why isn't it the same as an arbitrary point when I can increase it like this?

Ps. Would you also know where I can find more of these problems? Or perhaps what these are even called?

 

[haruspex]

I'm increasing the distance a by some dx

Not that I can see. You increased x by dx, which is valid, but a is a constant, so you cannot increase that. Increase in a would be by da, and I only see one integration stage.
To get the force on the second rod, you need to find the force on each part of the rod. There will be two integration stages.
Consider an element dx at x from the start of the first rod and the field it generates at y from the start of the second rod.

 

[Vriska]

It's the field at the start of the second rod but I'm increasing the distance a by some dx and summing it till 2L+a

Not that I can see. You increased x by dx, which is valid, but a is a constant, so you cannot increase that. Increase in a would be by da, and I only see one integration stage.
To get the force on the second rod, you need to find the force on each part of the rod. There will be two integration stages.
Consider an element dx at x from the start of the first rod and the field it generates at y from the start of the second rod.

But that's just adding a + y in the denominator of the integral, it's equivalent to E'(x) > E(x+a). I'm getting the same result. After looking at the answer in my book I notice it has $$ log(\frac{[L]^2}{[L]^3}) $$ term. I think exponentials do not accept dimensional arguments. Am I right in my assessment?

Also please tell me what these problems are called if you know.

 

[haruspex]

But that's just adding a + y in the denominator of the integral,

Yes, for the first integral. But then you have a second integral to do.
Please post the detailed steps.

 

[Korak Biswas]

Field due to the first rod on an infinitesimal line element at a point $(x',0)$ will be
$$E(x') = \int_0^{L} \frac{k\lambda dx}{(x'-x)^2}$$

Force on the second rod due to the first one will be
$$F = \int_{L+a}^{2L+a} \lambda E(x') dx'$$

 

[Korak Biswas]

Field due to the first rod on an infinitesimal line element at a point $(x',0)$ will be
$$E(x') = \int_0^{L} \frac{k\lambda dx}{(x'-x)^2}$$

Force on the second rod due to the first one will be
$$F = \int_{L+a}^{2L+a} \lambda E(x') dx'$$