Find the force of interaction - 2

Electricity and Magnetism Level 3

Find the electrostatic force of interaction between two parallel non conducting wires each of length $l$ , charge density per unit length $\lambda$ , and separated by a distance $d$ .

Give your answer as the value of this force in $\text{ Newtons }$ using $\lambda = 10 \mu \text{C/ m}$ , $l = 1 \text{m}$ , $\text{ d = 1m }, \frac{1}{4 \pi \epsilon_{0} } = 9 \times 10^9\text{ N } m^2/ C^2$

Note :

The wires completely face each other (like if their equations were $0 \leq x \leq L, y,z= 0$ and $0 \leq x \leq L, y= d, z=0$ )

The answer is 0.746.

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Jatin Yadav
Apr 29, 2014

Consider two elements on the wires of thickness $dx$ and $dy$ respectively on the wires as shown:

Their charges are $\lambda dx$ and $\lambda dy$ respectively. Distance between these elements is clearly $\displaystyle \sqrt{(y-x)^2+d^2}$

Hence, the coloumbic force of intersection between the elements is:

$dF = \dfrac{\lambda^2 \text{ dx dy}}{4 \pi \epsilon_{0} ((y-x)^2+d^2)}$

Now, by symmetry, we know that the force would be vertical, so we take only $\sin \theta$ component.

$\sin \theta = \dfrac{d}{\sqrt{(y-x)^2+d^2}}$

$F_{\text{net}} = \displaystyle \int dF \sin \theta$

= $\displaystyle \int_{0}^{l} \int_{0}^{l} \dfrac{\lambda^2 d }{4 \pi \epsilon_{0} ((y-x)^2+d^2)^{3/2}} \text{dx dy}$

This was tagged calculus due to use of double integral. I leave it on you. Its simple.

Solving this, we get:

$\boxed{F = \dfrac{\lambda^2}{2 \pi \epsilon_{0} d} \bigg(\sqrt{l^2+d^2} - d \bigg)}$

You can verify it by using approximation $l <<< d$ , in which case it gives the force as $\dfrac{ (\lambda l)^2}{4 \pi \epsilon_{0} d^2}$ (Compare to $F = \dfrac{q_{1}q_{2}}{4 \pi \epsilon_{0} r^2}$ )

_Note : _ Anyone unable to solve the definite double integral can freely ask as comment.

Can someone explain where i went wrong? I considered a small element $dx$ on one rod, $x$ distance from one end, with charge $dq$ and integrated, ie, the field at that point due to the second rod times $dq$ (considering just the perpendicular component), and got $0.9$ as the answer...

Pratik Shastri - 7 years, 1 month ago

Check out my solution.

Pinak Wadikar - 7 years ago

Why did u take two unit particles diagonally apart

Samarth Rastogi - 7 years, 1 month ago

He choose them for the case of generality, You could have just as easily chosen a straight line.

Beakal Tiliksew - 7 years, 1 month ago

Dude i did tht and taking d charge as lambda × l..n applied d formula (k×(lambda×l)^2)/d^2

Samarth Rastogi - 7 years, 1 month ago

@Samarth Rastogi – Hi friend, here we are taking force between two small elements, not the whole wires. You cannot apply the point charge force equation to a distribution of charges. Note that in the expression(in box) if you use $l <<< d$ , you get the answer same as that given by coloumbic force between two charges. This is because if $l<<<d$ , the wire are so small that they can be considered point charges.

jatin yadav - 7 years, 1 month ago

@Jatin Yadav – True tht thnks man

Samarth Rastogi - 7 years, 1 month ago

@Samarth Rastogi – Try also this inspred problem

Beakal Tiliksew - 7 years, 1 month ago

Can someone please post a solution for the double integral?

A K - 7 years, 1 month ago

Sure.

First, you should know that $\displaystyle \int \dfrac{dz}{(a^2+z^2)^{3/2}} = \dfrac{1}{a^2} \dfrac{z}{\sqrt{z^2+a^2}}$

Here, if $a = d$ , Then the double integral is changed to:

$\dfrac{\lambda^2}{4 \pi \epsilon_{0} d} \displaystyle \int_{0}^{l} \bigg(\dfrac{l-x}{\sqrt{(l-x)^2+d^2}} + \dfrac{x}{\sqrt{x^2+d^2}}\bigg) dx$

Now, using $\displaystyle \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ ,

We get $\displaystyle \dfrac{\lambda^2}{2 \pi \epsilon_{0} d} \int_{0}^{l} \dfrac{x dx}{\sqrt{x^2+d^2}}$

= $\dfrac{\lambda^2}{2 \pi \epsilon_{0} d} \bigg(\sqrt{l^2+d^2} - d \bigg)$

jatin yadav - 7 years, 1 month ago

Nice one. Thanks, Jatin.

A K - 7 years, 1 month ago

Pinak Wadikar
May 5, 2014

We can solve the problem without using the double integral method by using the generalized formula for electric field by finite line of charge.

Force of interaction 2 solution

Hey, I did it in the same manner!

Shaan Vaidya - 7 years ago

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Calvin Lin Staff - 7 years ago

Find the force of interaction - 2

The answer is 0.746.

2 solutions

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