A parallel-plate capacitor is connected to a battery and stores 3.5 nC of charge. A dielectric sheet with relative permittivity k = 2 is inserted between the plates while the battery remains connected. By how much does the charge change?

• A. 3.5 nC
• B. 1.75 nC
• C. 7.0 nC
• D. 0 nC

Answer: (A)

When a capacitor stays connected to a battery, the voltage across it is fixed by the battery. Inserting a dielectric increases the capacitance by a factor k, so the stored charge, Q, increases proportionally: Q = C V, with V constant.

Here, the initial charge is 3.5 nC. After inserting the dielectric with k = 2, the capacitance doubles, so the final charge becomes Q_final = k times the initial amount = 2 × 3.5 nC = 7.0 nC. The question asks for how much the charge changes, which is the difference: ΔQ = Q_final − Q_initial = 7.0 nC − 3.5 nC = 3.5 nC.

So the charge increases by 3.5 nC. The final charge would be 7.0 nC, not the change; the 0 nC choice would imply no effect from the dielectric, which isn’t the case here.