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Step-by-Step Solution
Step 1: Identify the System and Initial Conditions
We have a parallel-plate capacitor of capacitance $C$ (which already factors in the dielectric constant $K$). It is charged to a potential $V$, so the charge on it is $Q = CV$. The initial energy stored in the capacitor is given by:
$U_{\text{initial}} = \frac{1}{2}\,C\,V^2$
Step 2: Removing the Dielectric
When the dielectric is removed, the capacitor is isolated (not connected to any battery or circuit). Hence, the charge $Q$ on the plates remains constant. However, the capacitance changes because the dielectric has been removed. Let $C_{\text{no dielectric}}$ be the new capacitance (smaller than $C$), but the charge stays fixed at $Q$.
Step 3: Energy After Removing the Dielectric
After removing the dielectric, the voltage across the plates becomes $V_{\text{new}} = \frac{Q}{C_{\text{no dielectric}}}$. The new energy of the capacitor can be written as:
$U_{\text{after removal}} = \frac{Q^2}{2\,C_{\text{no dielectric}}}$
Since $Q = CV$ (the original charge), we can compare $U_{\text{after removal}}$ with $U_{\text{initial}}$. Detailed calculations would show that, under a slow (quasi-static) process of removing the dielectric, no net energy is lost or gained by the system. Any minor work done in changing the plate separation or reintroducing the dielectric is returned in the re-insertion process.
Step 4: Reinserting the Dielectric
Placing the dielectric back into the capacitor again changes its capacitance back to $C$. The voltage across the capacitor adjusts accordingly, but the charge is still the same $Q$. The final energy returns to the initial form:
$U_{\text{final}} = \frac{1}{2}\,C\,V^2$
Step 5: Conclude the Net Work Done
Because the capacitor is isolated throughout the process, there is no external energy exchange that remains in the system. After the entire cycle (removing and then reinserting the dielectric), the system’s energy returns to its original value:
$U_{\text{final}} - U_{\text{initial}} = 0$
Hence, the net work done by or on the system in this process is zero.
Final Answer
The correct answer is: zero.
