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You're essentially asking to compute $\nabla\cdot\boldsymbol{\sigma}$ in coordinate-free terms. Here's a possible way, although it requires some tricks, in particular the identities $$\text{tr}(\nabla \mathbf{a})=\text{tr}(\nabla \mathbf{a}^\mathsf{T})=\nabla\cdot \mathbf{a}$$ $$\nabla\cdot\nabla \mathbf{a}=\Delta \mathbf{a}$$ ...
You just made some math mistakes. You made a mistake when you did $Q = h\int_A kr$. You got $Q = h\pi k r^2$, but you should have gotten $Q = \frac{2}{3} h\pi k r^3$. Notice how this second expression has units of charge while the first one doesn't. Another mistake you make is that you say $\frac{1}{r} \frac{\partial rE(r)}{\partial r} = \frac{1}{r} ... 2 Actually the COM for the 2-body problem is the essential feature in this subject and with respect to it, both the Earth and the Sun rotate. Indeed, motion is relative, the relativity of it is even easier to understand in the Galilean Relativity than in Special Relativity. The Heliocentric view is actually the correct opinion that the Sun of our planetary ... 1 The numbers of deegres of freedom for a planar body in the plane is $$\text{2 (coordinates of CM)}+1\text {(angles to determine the body orientation)}=3.$$ As you correctly recognize, the constraint is one: $$|\vec {OA}|=R,$$ so you need two numbers$\varphi, \theta$to localize the body in the plane. So, you have the first,$\varphi$, and I gave you a big ... 1 Disclaimer: I updated my previous answer as I found a more accurate formula. The rotation angle$\theta\$ is the angle between the vernal point and the meridian at Greenwich. This also corresponds with the Sidereal Time at Greenwich, converted to radians. For a given UTC time and a given date, the corresponding Greenwich (Mean) Sidereal Time (in hours and ...