Powering-invariant and normal not implies quotient-powering-invariant

Statement
It is possible to have a group $$G$$ and a powering-invariant normal subgroup $$H$$ of $$G$$ that is not quotient-powering-invariant: in other words, $$H$$ is a normal subgroup of $$G$$ that is also a powering-invariant subgroup of $$G$$, but there exists a prime number $$p$$ such that $$G$$ is $$p$$-powered and $$G/H$$ is not $$p$$-powered.

We can in fact construct our example such that both $$G$$ and $$H$$ are rationally powered groups and $$G/H$$ contains a divisible group as subgroup where every element has infinitely many $$n^{th}$$ roots for all $$n > 1$$.