Charged Crux

We have to know that the Electric Field strength at a distance $r$ from a wire is: $E=\frac{\lambda }{2 \pi \varepsilon_0}\cdot \frac{1}{r}$

We also have know that: $F=E \cdot q$

To solve this problem we will first take the first wire as fixed and we will find all the forces acting on the second wire (meaning forces due to the electric field of the first wire). To find the solution we have to add only the components of those forces that are perpendicular to the second wire.

A scheme will be helpful to understand:

Useful image for understanding: there is a Cartesian plane and a red spot on the y axes representing a wire

We can see a Cartesian plane. The second wire lies along the $x$ -axis, while the red spot on the $y$ -axis represents the first wire, that is perpendicular to the plane.

Firstly we have to find the force on every arbitrarily small part of the second wire due to the electric field of the first wire. We will call $\delta x$ an arbitrarily small part of the second wire (painted in purple in the scheme) and $x$ will be its distance from the origin. $h$ will be the distance between the wires. The distance between $\delta x$ and the first wire is: $\sqrt{h^2+x^2}$ Therefore the electric field in $\delta x$ is: $\frac{\lambda }{2 \pi \varepsilon_0}\cdot \frac{1}{\sqrt{h^2+x^2}}$ The charge in $\delta x$ is: $\lambda \cdot \delta x$ Therefore the force acting on $\delta x$ is: $\frac{\lambda ^2}{2 \pi \varepsilon_0}\cdot \frac{1}{\sqrt{h^2+x^2}} \delta x$

We need to find the component of this force that is perpendicular to the second wire. It is: $\frac{\lambda ^2}{2 \pi \varepsilon_0}\cdot \frac{1}{\sqrt{h^2+x^2}} \cdot \cos{\theta} \quad \delta x$ From the scheme we see that: $\cos{\theta}=\frac{ h }{ \sqrt{h^2+x^2} }$ Therefore the requested perpendicular component is (painted in yellow in the scheme): $F_{\perp}=\frac{\lambda ^2 h}{2 \pi \varepsilon_0}\cdot \frac{1}{h^2+x^2} \delta x$

Now we have to add up all these forces, and we can do it by evaluating the integral: $\int_{-\infty}^{+\infty} \frac{\lambda ^2 h}{2 \pi \varepsilon_0}\cdot \frac{1}{h^2+x^2} dx =$ $\frac{\lambda ^2 h}{2 \pi \varepsilon_0}\cdot \int_{-\infty}^{+\infty}\frac{1}{h^2+x^2} dx =$ $\frac{\lambda ^2 h}{2 \pi \varepsilon_0}\cdot \int_{-\infty}^{+\infty}\frac{1}{h^2+x^2} dx =$ $\frac{\lambda ^2 h}{2 \pi \varepsilon_0}\cdot \left.\frac{\tan^{-1}\left(\frac{x}{h}\right)}{h}\right|_{-\infty}^{+\infty} =$ $\frac{\lambda ^2 h}{2 \pi \varepsilon_0}\cdot \frac{\pi}{h} =\frac{\lambda ^2 }{2 \varepsilon_0}$

We can see that the result does not depend on the distance between the wires.

Substituting $8\cdot 10^{-6}$ we obtain as a result: $\fbox{3.6141}$

Consider $\hat n$ to be the common normal vector to both the wires, $a$ to be the distance between the two wires, and $\theta$ the angle which the radius vector, $\vec r$ , of an infinitesimal length $dx$ on the wire $2$ , makes with $\hat n$

We can say that the length $dx$ will be at a distance $x=a\tan\theta$ from the intersection of wire $2$ with $\hat n$ . Thus $dx=a\sec^2\theta d\theta$

The force due to wire $1$ on length $dx$ is $dF=\frac{\lambda}{2\pi\epsilon_0 r} \lambda a\sec^2 \theta\ d\theta$ We can observe that the horizontal component of this force(Considering $\hat n$ as vertical direction) will cancel out when considering the net force on the whole of wire $2$ , by symmetry. So we simply need to sum the vertical components. Thus $F_{total}=\int^{\theta=\pi}_{\theta=0} dF\cos\theta$ $F_{total}=\int^{\theta=\pi}_{\theta=0} \frac{\lambda}{2\pi\epsilon_0 r} \lambda a\sec^2 \theta\ d\theta \cos\theta$ $F_{total}=\int^{\theta=\pi}_{\theta=0} \frac{\lambda}{2\pi\epsilon_0 r\cos\theta} \lambda a\ d\theta$ Substituting $r\cos\theta=a$ , $F_{total}=\int^{\theta=\pi}_{\theta=0} \frac{\lambda}{2\pi\epsilon_0 a} \lambda a\ d\theta$ $F_{total}= \frac{\lambda^2}{2\epsilon_0 }$

One of the wires creates a field of the strength $\mathbf{E}=\frac{\lambda}{2\pi r\varepsilon_0}\hat{\mathbf{r}}$ . This field acts on the other wire with the net force of $\mathbf{F}=\int{\mathbf{E}\lambda\,\mathrm{d}x}$ , where $x$ axis is along the second wire.

For symmetry reasons only the component of the force which is orthogonal to both wires will remain. Substituting and taking into account that $F_z=F\frac a r$ , where $a$ is the distance between the wires, we get

$F_z = \frac{a\lambda^2}{2\pi\varepsilon_0}\int_{-\infty}^\infty{\frac{1}{a^2+x^2}\,\mathrm{d}x}=\frac{\tau^2}{2\varepsilon_0}=3.62\,\mathrm{N}$ .

Since, Electric field due to a uniformly charged rod at a point perpendicular to it at a distance 'r' midway is E = 2Kג/r So, force on an elemental length of the other rod (considering horizontal components, since vertical component of force cancels due to symmetry) dF = dq*E = 2Kגdq/r Since, dq = גdx so, dF = 2K\sqr{ג}dx/r Now, by Pythagoras theorem r = \sqrt{x^2+k^2} where 'x' is the distance of element from midpoint of rod and 'k' is the shortest distance between two rods So, total force on rod can be obtained by integrating dF by taking limits from -∞ to +∞

To visualize the two wires, let wire 1 run into the page, and wire 2 run horizontally on top of wire 1 when drawn on paper. We will find the force experienced by wire 2. This force will be given by $F= \Sigma (E_{perp} \Delta q)$ , where $\Delta q$ is a small charge element on wire 2, and $E_{perp}$ is the component of the electric field perpendicular to the wire (we know that there is no horizontal force due to the symmetry).

Now we will construct an infinitely long cuboid which extends horizontally, with wire 2 as one of its sides.The width of the cuboid is twice the separation of the wires (so that the cuboid is symmetrical about wire 1), and the breadth is an infinitesimal amount $\Delta w$ . We will apply Gauss's Law on this surface we constructed.

Now the electric flux through the surface of the cuboid will only be due to the top and bottom surfaces. This is given by $2\Sigma (E_{perp} \Delta A) = 2\Sigma (E_{perp} \Delta l \Delta w)$ . The charge enclosed is the charge of wire 1 enclosed in the cuboid, which is $\lambda\Delta w$ . Therefore, by Gauss' Law:

$2\Sigma (E_{perp} \Delta w \Delta l) =\frac{\lambda\Delta w}{\epsilon _0}$

Simplifying, $\Sigma (E_{perp} \Delta l) =\frac{\lambda}{2\epsilon _0}$

What we want is $F= \Sigma (E_{perp} \Delta q)$ , so we note that: $F= \Sigma (E_{perp} \Delta q)=\Sigma (E_{perp} \lambda \Delta l)$

Thus, we get $F=\lambda \Sigma (E_{perp} \Delta l)= \frac{\lambda ^2}{2\epsilon _0}$

Substituting the values, $F=\frac{(8\times10^{-6})^2}{2 \times (8.85 \times10^{-12})}=3.62 \mathrm{ N}$