Answer:
The expression of the field E as the point P becomes very far from the ring is:
[tex]\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x} \\\left \{ {{\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{1}{x^2}\vec{x} \mapsto x>0} \atop {\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{-1}{x^2}\vec{x} \mapsto x< 0 }} \right.[/tex]
Step-by-step explanation:
The Electric field expression is:
[tex]\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}[/tex]
To determine the asked expression we use limits. If we consider that x≫R, this is the same as considering the radius insignificant respect the x distance. Therefore we can considerate than from this distance X, the radius R tends to zero:
[tex]\displaystyle\lim_{R \to{}0}{\vec{E}(x)}=\lim_{R \to{}0}{\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}}\rightarrow\frac{q}{4\pi\epsilon_0} \frac{x}{(0^2+x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{x}{(x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{\cancel{x}}{|x|^{\cancel{3}}}\vec{x}=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x}[/tex]
The expression for the electric field of the ring as the point P becomes very far from the ring is [tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex].
Let suppose that the ring has an uniform linear electric density ([tex]\lambda[/tex]). A formula for the electric field at point P ([tex]E[/tex]) in rectangular coordinates is shown below:
[tex]\vec E = (E_{x}, E_{y}, E_{z})[/tex] (1)
Where:
Each component of the electric field are defined by the following integral formulae:
[tex]E_{x} = \int\limits^{2\pi}_{0} {\sin \theta \cdot \cos \phi} \, dE[/tex] (2)
[tex]E_{y} = \int\limits^{2\pi}_{0} {\sin \theta\cdot \sin\phi} \, dE[/tex] (3)
[tex]E_{z} = \int\limits^{2\pi}_{0} {\cos \theta} \, dE[/tex] (4)
Where:
By Coulomb's law and trigonometric and geometric relationships, we expand and solve each integral as following:
[tex]E_{x} = \frac{R}{\sqrt{x^{2}+R^{2}}}\int\limits^{2\pi}_{0} {\cos \phi} \, dE = \frac{k\cdot \lambda\cdot R^{2}}{(x^{2}+R^{2})^{3/2}}\int\limits^{2\pi}_{0} {\cos \phi} \, d\phi = 0[/tex]
[tex]E_{y} = \frac{R}{\sqrt{x^{2}+R^{2}}}\int\limits^{2\pi}_{0} {\sin \phi} \, dE = \frac{k\cdot \lambda\cdot R^{2}}{(x^{2}+R^{2})^{3/2}}\int\limits^{2\pi}_{0} {\sin \phi} \, d\phi = 0[/tex]
[tex]E_{z} = \frac{k\cdot \lambda\cdot x \cdot R}{(x^{2}+R^{2})^{3/2}} \int\limits^{2\pi}_{0}\, d\phi = \frac{x\cdot k \cdot (2\pi\cdot \lambda\cdot R)}{(x^{2}+R^{2})^{3/2}} = \frac{x\cdot k\cdot Q}{(x^{2}+R^{2})^{3/2}}[/tex] (5)
Where [tex]k[/tex] is the electrostatic constant.
If [tex]x >> R[/tex], (5) is simplified into the following expression:
[tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex] (6)
Where [tex]Q[/tex] is the electric charge of the entire ring.
Please notice that (6) tends to be zero when [tex]x \to \infty[/tex]. The expression for the electric field of the ring as the point P becomes very far from the ring is [tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex]. [tex]\blacksquare[/tex]
To learn more on electric fields, we kindly invite to check this verified question: https://brainly.com/question/12757739
