Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
while the voltage supply remained connected.
When the voltage supply remains connected, the potential difference across the capacitor remains constant at V₀. As the mica sheet (dielectric constant K = 6) is inserted, the capacitance increases to C = KC₀ = 6C₀. Since Q = CV and V is constant, the charge on the plates increases to Q = 6Q₀. The energy stored U = ½CV² also increases sixfold to U = 6U₀. The additional charge flows from the battery onto the plates.
after the supply was disconnected.
After the supply is disconnected, the charge remains constant at Q₀ as there is no path for charge to flow. When the mica sheet is inserted, the capacitance increases to C = KC₀ = 6C₀. Since V = Q/C and Q is constant, the potential difference decreases to V = V₀/6. The energy stored U = ½Q²/C decreases to one-sixth of its original value, U = U₀/6. The decrease in energy is accounted for by the work done in inserting the dielectric slab against the attractive forces.
Explanation
This question tests understanding of how dielectrics affect capacitors under two different conditions. In case (a), the battery maintains constant voltage, so the capacitor draws more charge. In case (b), with no battery connected, charge is conserved. The key formulas from the textbook are: C = KC₀ (capacitance increases by factor K), and for linear dielectrics, the potential difference V = V₀/K when charge is constant. The energy changes follow from U = ½CV² or U = ½Q²/C depending on which quantity remains constant.