Question

At what distance along the z-axis is the electric field strength a maximum?

Answer 1

The electric field strength is a maximum at 2.0 cm along the z-axis.

Step-by-step explanation:

The electric field strength is a maximum at a distance of 2.0 cm along the z-axis. This can be determined using Coulomb's law, which states that the electric field strength decreases with distance from a charged object. As you move further away from the object along the z-axis, the electric field strength decreases.

Answer 2

The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is

dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2)

where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis.

However, cos(T) = x/sqrt(x^2 + R^2)

so the equation becomes

dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2)

dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5

Integrating around the ring you get

E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5

E = (R/2*e0)*x*(x^2 + R^2)^-1.5

we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is

dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x}

dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5}

to find the maxima set this = 0, giving

(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0

mult both side by (x^2 + R^2)^2.5 to get

(x^2 + R^2) - 3*x^2 = 0

-2*x^2 + R^2 = 0

-2*x^2 = -R^2

x = (+/-)R/sqrt(2)