Question

Find the expression for the electric field, E [infinity] , of the ring as the point P becomes very far from the ring ( x ≫ R ) in terms of the radius R , the distance x , the total charge on the ring q , and the constant k = 1 / ( 4 π ϵ 0 ).

Answer 1

Answer:

The expression of the field E as the point P becomes very far from the ring is:

$\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x} \\\left \{ {{\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{1}{x^2}\vec{x} \mapsto x>0} \atop {\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{-1}{x^2}\vec{x} \mapsto x< 0 }} \right.$

Step-by-step explanation:

The Electric field expression is:

$\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}$

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:

$\displaystyle\lim_{R \to{}0}{\vec{E}(x)}=\lim_{R \to{}0}{\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}}\rightarrow\frac{q}{4\pi\epsilon_0} \frac{x}{(0^2+x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{x}{(x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{\cancel{x}}{|x|^{\cancel{3}}}\vec{x}=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x}$