To differentiate, let E=qxrho^(-3/2), "where " rho = a^2+x^2 E=qxρ−32,where ρ=a2+x2
By the product rule:
E' = q (rho^(-3/2) + x (-3/2 rho^(-5/2) rho')) and rho' = 2x
= q rho^(-3/2) (1 - 3 x^2 /rho )
As rho^(-3/2) ne 0 we look at the other solution:
E' = 0 implies 3x^2 = a^2+x^2 implies x = pm a/sqrt2
We can reality check this. The field at the dead centre will be zero due to symmetry. For small x, the relationship is linear: E approx q/a^3 x
And the field at a distance will be the usual inverse square, as the disc tends to a point source and the field falls away.
The sign on E indicates a force acting away from the origin on a positive test charge, so in terms of magnitude (absE), it also follows that these critical points are maxima, ie no need to compute the second derivative.
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