We will test each option to see if it can be true or not. Links to diagrams are provided.
\[\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\]
Let circle $Q$ be inside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR = p$, and $QR = q$, so $PR - QR = PQ = p-q.$

\[\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\]
Let circle $Q$ be outside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR = p$, and $QR = q$, so $PR + QR = PQ = p+q.$

\[\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\]
Let circle $Q$ be outside circle $P$ and not tangent to circle $P$, and the intersection of $\overline{PQ}$ with the circles be $R$ and $S$ respectively. $PR = p$ and $QS = q$, and $PR + QS < PQ$, so $p+q < PQ.$

\[\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\]
Let circle $Q$ be inside circle $P$ and not tangent to circle $P$, and the intersection of $\overline{PQ}$ with the circles be $R$ and $S$ as shown in the diagram. $PR = p$ and $QS = q$, and $QS < QR$, so $PR - QS < PR - QR$, and $PR - QR = PQ$, so $p-q < PQ.$

Since options A, B, C, and D can be true, the answer must be $\boxed{E}$.
Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q$. Draw $PQ$. Which of the following statements is false? 
$\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\\  \textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\\  \textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\\  \textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\\ \textbf{(E)}\ \text{none of these}$