A dielectric material is placed between the plates of a capacitor while the battery is still connected. What happens to the capacitance (C), voltage (V) and charge (Q)?

Question

BlackTom AI · Accepted Answer

To tackle this question, I’ll walk through what happens to each quantity when a dielectric is placed between the plates while a battery remains connected.
Option a: C increases, V decreases, Q remains the same. This is inconsistent with a battery-connected scenario: the voltage across the capacitor remains fixed by the battery, so V cannot decrease. Also, while C would increase due to the dielectric, Q would not stay the same because Q = C × V and V is constant, so Q must increase.
Option b: C decreases, V increases, Q remains the same. This contradicts the physics here: inserting a dielectric increases the capacitor’s ability to store charge, so C increases, not decreases. Since the battery fixes V, V does not increase either. Thus this optionmisstates both C and V, and keeping Q the same is incompatible with C changing while V is fixed.
Option c: C increases, V remains the same, Q increases. This aligns with the correct reasoning: a dielectric raises the capacitance (C ↑) when connected to a fixed voltage source, the voltage V stays equal to the battery (V stays the same), and the charge Q increases because Q = C × V and C has increased while V is unchanged.
Option d: C decreases, V remains the same, Q decreases. This is opposite to what a dielectric does to capacitance (C does not decrease when a dielectric is placed between the plates with the battery connected). Since C would increase, not decrease, and Q = C × V with V fixed would also increase, this option is incorrect.
In summary, the only option that consistently describes the physical behavior under these conditions is the one where C increases, V stays the same, and Q increases.

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