# Question #ddefc

May 5, 2017

To differentiate, let $E = q x {\rho}^{- \frac{3}{2}} , \text{where } \rho = {a}^{2} + {x}^{2}$

By the product rule:

$E ' = q \left({\rho}^{- \frac{3}{2}} + x \left(- \frac{3}{2} {\rho}^{- \frac{5}{2}} \rho '\right)\right)$ and $\rho ' = 2 x$

$= q {\rho}^{- \frac{3}{2}} \left(1 - 3 {x}^{2} / \rho\right)$

As ${\rho}^{- \frac{3}{2}} \ne 0$ we look at the other solution:

$E ' = 0 \implies 3 {x}^{2} = {a}^{2} + {x}^{2} \implies x = \pm \frac{a}{\sqrt{2}}$

We can reality check this. The field at the dead centre will be zero due to symmetry. For small $x$, the relationship is linear: $E \approx \frac{q}{a} ^ 3 x$

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 ($\left\mid E \right\mid$), it also follows that these critical points are maxima, ie no need to compute the second derivative.