Potential energy stored in the capacitor is U=1/2CV^2. Dielectric with d=4 will increase the capacitance by 4. Calculate the energy stored in the capacitor before and after the dielectric is inserted - the difference will be the work done to insert the dielectric.
The justification why ΔU is the work done in this case is somewhat fishy, at least from what I can think of at the moment. There is another example, where you have two connected capacitors with no battery and a dielectric is inserted in one of them causing change of charge. Calculating the total energy in both capacitors before and after the insertion shows that there is a change in energy. In that case it's easy to argue that the change of energy comes from work because there are no other sources - the two capacitors are a closed system. In your problem there is a transfer of energy from the battery, so...
What's the source of the problem?
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