= Solution
Use $Q=CV$ for the magnitude of charge on either plate. The fully filled <capacitance> is $C_f=2\pi\epsilon_0\epsilon L/\ln(b/a)$, while the initial value is given in part (c). At the same fixed <electric potential difference>,
$$
\boxed{\Delta Q=V(C_f-C_i)
=\frac{2\pi\epsilon_0(\epsilon-1)V(L-x)}{\ln(b/a)}.}
$$
For $\epsilon>1$ and positive $V$, this is an increase in the positive plate's <electric charge> magnitude, supplied by the voltage source; the negative plate acquires an equal opposite change. The net free charge of the two-plate capacitor remains zero. Thus “charge on the capacitor” has its usual circuit meaning, not the algebraic sum over both plates.
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