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{\bf Solution Paul Trap} \\%
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\bigskip \noindent

In the quasi-static approximation the field is the electrostatic field 
with the given boundary conditions at the time in question. Due to the 
cylindrical symmetry we have $\Phi=\Phi(\rho,z;t)$. To get the boundary 
conditions on the end electrodes right, we set
$$ \Phi\ =\ \left(z^2-\frac{1}{2}\,\rho^2-d^2\right)\,F(\rho,z,t)\,. $$
The boundary condition on the ring electrode implies 
$\rho^2/2=z^2+d^2/2$ and on this boundary
$$ \Phi\ =\ V_0\,\sin(\omega t)\ =\ -\frac{3}{2}\,d^2\,F(\rho,z,t)\,.$$
Therefore, the function $F$ depends only on $t$ 
$$ F(t)\ =\ - \frac{2}{3}\,\frac{V_0}{d^2}\,\sin(\omega t)\,,$$
so that the potential becomes
$$ \Phi\ =\ - \frac{2}{3}\,\frac{V_0}{d^2}\,
   \left(z^2-\frac{1}{2}\,\rho^2-d^2\right)\,\sin(\omega t)\,, $$
which is now easily seen to be the solution:
$$ \nabla^2\Phi = - \frac{2}{3}\frac{V_0}{d^2}\,\sin(\omega t)
   \,\nabla^2 \left(z^2-\frac{1}{2}\,\rho^2-d^2\right)\ =\ 0\,.$$
The electric field is given by
$$ \vec{E}\ = - \nabla\cdot\Phi\ =\ -\left(\hat{\rho}\,
                \frac{\partial\Phi}{\partial\rho} 
              + \hat{z}\,\frac{\partial\Phi}{\partial z}\right)\
   =\ \frac{2}{3}\,\frac{V_0}{d^2}\,\left(2\,z\,\hat{z}
              -\rho\,\hat{\rho}\right)\,\sin(\omega t)\,. $$

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