Respuesta :
Answer:
[tex]y_{\rm max} = \frac{R}{\sqrt{2}}[/tex]
Explanation:
We should first calculate the electric field of a ring at a point along its axis.
We will use the following electric field formula for a point charge.
[tex]\vec{E} = k\frac{q}{r^2}\^r[/tex]
Since this formula is valid for point charges, we will separate our ring into infinitely many points, apply this formula to each of the points, and then add them up. Of course, this will take a lot of time, so we will use the method of integration.
First, we will choose an infinitesimal displacement, ds, whose charge is dq.
The electric field of the infinitesimal portion is
[tex]dE = k\frac{dq}{r^2}[/tex]
where dq can be written in terms of Q and R, and ds = Rdθ.
[tex]\frac{Q}{2\pi R} = \frac{dq}{Rd\theta}\\dq = \frac{Qd\theta}{2\pi}[/tex]
The distance from that point to an arbitrary point at the axis is
[tex]r = \sqrt{R^2 + y^2}[/tex].
Therefore, dE becomes
[tex]dE = k\frac{Qd\theta}{2\pi (R^2+y^2)}[/tex]
Since electric field is a vector, we have to separate the x- and y-components of dE, before integrate it. By symmetry, the x-components cancel out each other. So, we only calculate the y-components.
[tex]dE_y = dE\sin(\alpha) = dE\frac{y}{\sqrt{R^2 + y^2}}[/tex]
Combining these results,
[tex]dE_y = \frac{kQ}{2\pi}\frac{y}{(R^2+y^2)^{3/2}}d\theta[/tex]
Now, we can integrate over the ring
[tex]E_y = \int dE_y = \frac{kQ}{2\pi}\frac{y}{(R^2+y^2)^{3/2}} \int\limits^{2\pi}_0 {} \, d\theta = \frac{kQ}{2\pi}\frac{y}{(R^2+y^2)^{3/2}}2\pi\\\vec{E}_y = \frac{kQy}{(R^2+y^2)^{3/2}}\^y[/tex]
Now, we should find the maximum value of this function. In order to find this, we should look for the value where its derivative is equal to zero.
[tex]\frac{dE_y}{dy} = 0\\\frac{kQ}{(R^2+y^2)^{3/2}} - \frac{3kQy^2}{(R^2+y^2)^{5/2}} = 0\\\frac{kQ}{(R^2+y^2)^{3/2}} = \frac{3kQy^2}{(R^2+y^2)^{5/2}}\\1 = \frac{3y^2}{R^2+y^2}\\R^2+y^2 = 3y^2\\R^2 = 2y^2\\y = R/\sqrt{2}[/tex]
