Suppose $ \cos x \equal{} 0$ and $ \cos (x \plus{} z) \equal{} 1/2$. What is the smallest possible positive value of $ z$?

$ \textbf{(A) } \frac {\pi}{6}\qquad \textbf{(B) } \frac {\pi}{3}\qquad \textbf{(C) } \frac {\pi}{2}\qquad \textbf{(D) } \frac {5\pi}{6}\qquad \textbf{(E) } \frac {7\pi}{6}$
1. Given that $\cos x = 0$, we know that $x$ must be an odd multiple of $\frac{\pi}{2}$:
   \[
   x = \frac{\pi}{2} + k\pi \quad \text{for integer } k.
   \]

2. We are also given that $\cos(x + z) = \frac{1}{2}$. The cosine function equals $\frac{1}{2}$ at angles of $\frac{\pi}{3} + 2m\pi$ and $- \frac{\pi}{3} + 2m\pi$ for integer $m$. Therefore, we have:
   \[
   x + z = \frac{\pi}{3} + 2m\pi \quad \text{or} \quad x + z = -\frac{\pi}{3} + 2m\pi.
   \]

3. Substitute $x = \frac{\pi}{2} + k\pi$ into the equations:
   \[
   \frac{\pi}{2} + k\pi + z = \frac{\pi}{3} + 2m\pi \quad \text{or} \quad \frac{\pi}{2} + k\pi + z = -\frac{\pi}{3} + 2m\pi.
   \]

4. Solve for $z$ in each case:
   \[
   z = \frac{\pi}{3} + 2m\pi - \frac{\pi}{2} - k\pi = \frac{\pi}{3} - \frac{\pi}{2} + (2m - k)\pi = -\frac{\pi}{6} + (2m - k)\pi,
   \]
   \[
   z = -\frac{\pi}{3} + 2m\pi - \frac{\pi}{2} - k\pi = -\frac{\pi}{3} - \frac{\pi}{2} + (2m - k)\pi = -\frac{5\pi}{6} + (2m - k)\pi.
   \]

5. We need the smallest positive value of $z$. Consider the first equation:
   \[
   z = -\frac{\pi}{6} + (2m - k)\pi.
   \]
   To make $z$ positive, we need:
   \[
   -\frac{\pi}{6} + (2m - k)\pi > 0 \implies (2m - k)\pi > \frac{\pi}{6} \implies 2m - k > \frac{1}{6}.
   \]
   The smallest integer value for $2m - k$ that satisfies this inequality is $1$:
   \[
   2m - k = 1 \implies z = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}.
   \]

6. Now consider the second equation:
   \[
   z = -\frac{5\pi}{6} + (2m - k)\pi.
   \]
   To make $z$ positive, we need:
   \[
   -\frac{5\pi}{6} + (2m - k)\pi > 0 \implies (2m - k)\pi > \frac{5\pi}{6} \implies 2m - k > \frac{5}{6}.
   \]
   The smallest integer value for $2m - k$ that satisfies this inequality is $1$:
   \[
   2m - k = 1 \implies z = -\frac{5\pi}{6} + \pi = \frac{\pi}{6}.
   \]

7. Comparing the two possible values of $z$, $\frac{5\pi}{6}$ and $\frac{\pi}{6}$, the smallest positive value is $\frac{\pi}{6}$.

The final answer is $\boxed{\frac{\pi}{6}}$.