\section{Quad Misalignment}
We estimate the error in our determination of the efield correction in the event
that all inner and outer plates on each of the eight quads are displaced radially by a fixed amount,
and if all of the inner and outer plates share a common voltage error.
Suppose the plates are displaced by an amount $\Delta r$. The magic momentum will no longer be along the quad axis.
The on axis momentum will be shifted from $R_{magic}$ to $R_{md}+\Delta r$ corresponding to a change in momentum with respect to $p_m$ of
$\Delta p/p = \Delta r/R_{md}$. The radius of the closed orbit of the magic momentum muon will shift to 
$$R_{md}=R_0 +\Delta r- \eta \frac{\Delta p}{p} = R_0 +\Delta r- \frac{R_0}{1-n}\frac{\Delta r}{R_0} = R_0-\Delta r\frac{n}
{1-n}$$ so that $x_{md} = -\Delta r\frac{n}{1-n}$.

Recall that $$C_e = -2\frac{(\Delta p/p)\beta E_\rho}{B}$$
Now we have that
$$x = \eta\Delta p/p + x_{md} $$
$$\Delta p/p = \frac{x-x_{md}}{\eta}$$
where $x$ remains the distance from the unperturbed magic radius $R_0$. 
The dependence of electric field on radial displacement becomes
$$E_\rho \rightarrow n\left(\frac{\beta B}{R_0}\right)(x -\Delta r)$$
That means that
$$C_e \rightarrow -\frac{2}{B}\frac{(x-x_{md})}{\eta}\beta n\left(\frac{\beta B}{R_0}\right)(x-\Delta r)$$
$$C_e\rightarrow -2(1-n)n\frac{(x-x_{md})(x-\Delta r)}{R_0^2}$$
and
$$\vev{C_e} = -2(1-n)n\frac{\vev{x^2}-\vev{x}(x_{md}+\Delta r)+x_{md}\Delta r}{R_0^2}$$
The change in $C_e$ due to the quad offset is
\begin{eqnarray*}
\delta\vev{C_e}&=& -2(1-n)n\frac{-\vev{x}(x_{md}+\Delta r)+x_{md}\Delta r}{R_0^2}\\
&=& -2(1-n)n\frac{-\vev{x}\Delta r(-\frac{n}{1-n}+1)-\frac{n}{1-n}(\Delta r)^2}{R_0^2}\\
&=& -2(1-n)n\frac{-\vev{x}\Delta r\frac{1-2n}{1-n}-\frac{n}{1-n}(\Delta r)^2}{R_0^2}\\
\end{eqnarray*}
If $\Delta r=1$ mm and $\vev{x}=6$ mm  and $n=0.1$ then
 $$\delta C_e = 19.4 {\rm ppb}$$
and if $\vev{x}=0$ then $\delta C_e = 0.4$ ppb.
\section*{Quad Voltage Error}
The voltage depends quadratically along the radial axis as
$$V(x) = V_p\frac{x^2}{a^2}$$ where $a=0.05$m is the displacement of the plates from the center of the good field region and
$V_p$ the voltage on the plates.
The electric field along the radial axis is 
$$E_\rho=2V_p\frac{x}{a^2}$$
Suppose that there is a voltage error on the inner and outer quad plates. The 'worst' case scenario is
$\Delta V_{inner} = -\Delta V_{outer} = \Delta V$.
If $V_{inner}\rightarrow V_{inner}+\Delta V$ and $V_{outer}\rightarrow V_{outer}-\Delta V$ then
$$E_\rho\rightarrow 2V_p\frac{x}{a^2} + \frac{2\Delta V}{2a}$$
Then $$E_\rho = 0\rightarrow x_V=a\frac{\Delta V}{2V_p}$$ where $x_V$ is the displacement of the zero of the field from
the magic radius. Then
$$E_\rho(x) \rightarrow n\left(\frac{\beta B}{R_0}\right)(x - x_V)$$
Now it looks equivalent to a misalignment by $x_V$.
The closed orbit of the magic momentum shifts as before to 
 $$x_{mV} = -x_V\frac{n}{1-n}.$$
The change in $C_e$ due to the quad voltage error is
\begin{eqnarray*}
\delta\vev{C_e}&=& -2(1-n)n\frac{-\vev{x}(x_{mV}+x_V)+x_{md}x_V}{R_0^2}\\
&=& -2(1-n)n\frac{-\vev{x}x_V(-\frac{n}{1-n}+1)-\frac{n}{1-n}x_V^2}{R_0^2}\\
&=& -2(1-n)n\frac{-\vev{x}x_V\frac{1-2n}{1-n}-\frac{n}{1-n}x_V^2}{R_0^2}\\
\end{eqnarray*}
If $\Delta V/V=5$\% then $x_V= 1.25$ mm. If $\vev{x}=6$ mm  and $n=0.1$ then
 $$\delta C_e = 24.3 {\rm ppb}.$$
If we manage to zero $\vev{x}$ then $\Delta C_e = 0.6$ ppb.

\section*{Misalignment and voltage error}
IF we have both voltage errors and misalignments then the effects with add. 

%As in the case of misalignment, there is no first order effect on dispersion and

%$$\vev{C_e} = 2(1-n)n\frac{\vev{(x-x_m)(x- x_V)}{R_0^2} = $$
%If $\vev{x_e}=6$ mm, and $\Delta V/V = 0.05$ then $x_V=1.25$ mm and $\delta C_e=26.7$ ppb. 
