# Why would a rotating charged sphere not have time varying electric field?

I have seen Gauss Law being used for a uniformly charged hollow sphere rotating with $\omega$. How is that valid to use Gauss law since it is an electrostatic law and if it is valid, why do we get a net constant electric field outside the sphere inspite of the accelerating charges - does the net time dependence cancel out ? How do I see this cancellation intuitively ? Also, can I see intuitively why the net radiation emitted is zero instead of integrating the Poynting flux ?

• What are the assumptions on the sphere? Is the charge uniformly and continuously distributed across the surface? Aug 26, 2014 at 7:34
• yes, it is a uniform distribution; it could be a hollow sphere in which case, it is a surface distribution and as a special case, it is also true then for the solid sphere with a uniform volume charge distribution
– user56199
Aug 26, 2014 at 8:33

Take a look at the conventional form of Maxwell equations. They tell us that Gauss's law actually applies every time. However, to get the field $\vec{E}$ from the charge distribution by the usual methods, we also need to know that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$$ Because otherwise the field could not be generated by the electrostatic potential. We know that the magnetic field is always divergenceless and $$\nabla \times \vec{B} = \mu_0 (\vec{j} + \varepsilon_0\frac{\partial \vec{E}}{\partial t})$$ So if the charge distribution is fixed (not to change $\vec{E}$) and the current also, then we have no changes in the magnetic field $\vec{B}$, $\partial_t \vec{B}$ is zero and we can use our usual electrostatic methods to solve the problem.
EDIT: Before someone smacks me, the last statement has obviously some exceptions if we consider different initial and boundary conditions. A planar electromagnetic wave could be passing by and scattering off the sphere etc. I implicitly assume there are no $\partial_t \vec{B},\partial_t \vec{E}$ to begin with and that the boundary conditions (the boundary could be at finite or infinite distance) do not change in time.