= Solution
An element at $a(\cos\phi,\sin\phi,0)$ carries charge $dq=\lambda_0a\sin\phi\,d\phi$. Its field at the origin points opposite its radial position. By <Coulomb's law>,
$$
\mathbf E(0)=-\frac{\lambda_0}{4\pi\epsilon_0a}
\int_0^{2\pi}\sin\phi\big(\cos\phi\,\hat{\mathbf x}+\sin\phi\,\hat{\mathbf y}\big)\,d\phi
=-\frac{\lambda_0}{4\epsilon_0a}\hat{\mathbf y}.
$$
Consequently
$$
\boxed{\mathbf a(0)=-\frac{q\lambda_0}{4\epsilon_0ma}\hat{\mathbf y}.}
$$
For $q\lambda_0>0$ the particle starts along $-y$; for the opposite sign it starts along $+y$. Reflection symmetries make the $y$ axis invariant for motion released there from rest. The ring has zero net charge but a nonzero dipole moment, so zero total charge does not imply zero field at its centre.
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