Does the Ampère-Maxwell law fail for the field of a uniformly moving point charge?
I think I found a terrible contradiction in the Ampère-Maxwell law (the 4th <PERSON> equation) that really baffles me.
If a point charge q is in motion in vacuo with constant velocity $ \vec v $, it causes an electric field $ \vec E $ as well as a magnetic field $ \vec B $ at a point P. The vector-position $ \vec r $ of the point P with respect to the charge, the charge itself and $ \vec v $ determine a plane. We can imagine the charge moving along the x-axis of a cartesian plane, its equation of motion given by $ x = x_0 + vt $, $ v = \dfrac {dx}{dt} $ = const. 903
while the point P is located at a permanent position on the y-axis, y(P) = const.
The electric field caused by the charge q at P is given by
$$ \vec E(P) = \dfrac{q}{4 \pi \epsilon_0} \dfrac{1 - v^2/c^2}{[1 - (v^2/c^2) sin^2 \phi]^{3/2}} \dfrac{1}{r^2} \hat r $$
and the magnetic field is given by
$$ \vec B(P) = \dfrac{\mu_0 q}{4 \pi} \dfrac{1 - v^2/c^2}{[1 - (v^2/c^2) sin^2 \phi]^{3/2}} \dfrac{\vec v \times \hat r}{r^2} $$
So the electric field, being radial to the charge, is contained in the x0y plane, while the magnetic field is perpendicular to the same.
The Ampère-Maxwell law states that $$ \operatorname{curl} \vec B = \epsilon_0 \mu_0 \dfrac{\partial \vec E}{\partial t} $$
And we have
$$ \vec B = \epsilon_0 \mu_0 \vec v \times \vec E = \dfrac{\vec v \times \vec E}{c^2} $$
$$ \vec v \times \vec E = vE_y \hat z $$
$$ B_x = 0 $$ $$ B_y = 0 $$ $$ B_z = \dfrac{v}{c^2} E_y $$
Therefore $$ \operatorname{curl} \vec B = \dfrac{v}{c^2} \left[ \dfrac{\partial E_y}{\partial y} \hat x - \dfrac{\partial E_y}{\partial E_x} \hat y \right] $$
Calculating the derivative of $ \vec E $ with respect to time, we get
$$ \dfrac{dE_x}{dt} = \dfrac{qv}{4 \pi \epsilon_0} \left( 1 - \dfrac{v^2}{c^2} \right) \left[ x^2 + y^2 \left( 1 - \dfrac{v^2}{c^2} \right) \right]^\dfrac{-3}{2} \left[ 3x^2 \left[ x^2 + y^2 \left( 1 - \dfrac{v^2}{c^2} \right) \right]^{-1} -1 \right] $$
$$ \dfrac{dE_y}{dt} = - \dfrac{qv}{4 \pi \epsilon_0} \left( 1 - \dfrac{v^2}{c^2} \right) 3xy \left[ x^2 + y^2 \left( 1 - \dfrac{v^2}{c^2} \right) \right]^\dfrac{-5}{2} $$
Then, when I calculate $ \dfrac{\partial E_y}{\partial y} $ and $ \dfrac{\partial E_y}{\partial x} $ to insert in the above formula for $ \operatorname{curl} \vec B $, I do not obtain an equality to my surprise, which indicates that the law of Ampère-Maxwell may not be valid in this case.
So the following three propositions cannot be all true:
1. The formulas given above to calculate the electric and magnetic field at point P are correct.
2. The derivatives above were calculated correctly.
3. 903