Question

The electric field at the center of a ring of charge is zero. At very large distances from the center of the ring along the ring's axis the electric field goes to zero. Find the distance from the center of the ring along the axis (perpendicular to the plane containing the ring) at which the magnitude of the electric field is a maximum. The radius of the ring is 6.58 cm and the total charge on the ring is 8.87E-6 C.

Answer 1

Answer:

z = 1.16m

Explanation:

The electric field in a point of the axis of a charged ring, and perpendicular to the plane of the ring is given by:

$E_z=\frac{Qz}{(z^2+r^2)^{\frac{3}{2}}}$

z: distance to the plane of the ring

r: radius of the ring

Q: charge of the ring

you have that:

E_{z->0} = 0

E_{z->∞} = 0

To find the value of z that maximizes E you use the derivative respect to z, and equals it to zero:

$\frac{dE_z}{dz}=Q[\frac{1}{(z^2+r^2)^{3/2}}+z(-\frac{3}{2})\frac{1}{(z^2+r^2)^{5/2}}(2z)]=0\\\\(z^2+r^2)^{5/2}=3z^2(z^2+r^2)^{3/2}\\\\(z^2+r^2)^2=3z^4\\\\z^4+2z^2r^2+r^4=3z^4\\\\2z^4-2z^2r^2-r^4=0\\\\z^2_{1,2}=\frac{-(-2)+-\sqrt{4-4(2)(-1)}}{2(2)}=\frac{2\pm 3.464}{4}$

you take the positive value:

$z^2=\frac{2+3.464}{4}=1.366\\\\z=1.16m$

hence, the distance in which the magnitude if the electric field is maximum is 1.16m