A capacitor is fully charged before the battery is disconnected. Then a dielectric material is placed between the plates of a capacitor. What happens to the capacitance (C), voltage (V) and charge (Q)?

Question

BlackTom AI · Accepted Answer

We are given a capacitor that is fully charged, then the battery is disconnecting, and a dielectric is inserted between the plates. We need to track how C, V, and Q change.

Option a: C increases, V remains the same, Q increases.
- Why this is unlikely: When the capacitor is isolated (battery disconnected), the charge Q on the plates cannot change because there is no path for charge to flow. This makes the claim 'Q increases' incorrect. Also, inserting a dielectric does increase C, but the voltage need not stay the same in an isolated scenario.

Option b: C increases, V decreases, Q remains the same.
- Why this is correct: Inserting a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant (C → C'), since C = ε0εrA/d. The charge remains fixed because the circuit is open after disconnection, so Q is conserved. The relationship Q = CV implies that if Q is constant and C increases, the voltage V must decrease. This aligns with the fundamental capacitor equation and the given conditions.

Option c: C decreases, V remains the same, Q decreases.
- Why this is incorrect: Inserting a dielectric does not decrease capacitance; it increases it. Also, Q cannot decrease when the plates are isolated, since there is nowhere for charge to leave the plates. The combination of these two wrong parts makes this option invalid.

Option d: C decreases, V increases, Q remains the same.
- Why this is incorrect: Again, the dielectric increases C, not decreases it. While Q remaining the same is true for an isolated capacitor, the predicted change in C and V is opposite to what actually happens: V would decrease when C increases (for fixed Q). Therefore this option misstates both the trend in C and the consequence for V.

In summary, the core ideas are: a dielectric increases the capacitance of an isolated capacitor, charge on the plates stays constant when the battery is disconnected, and since Q = CV, an increase in C with fixed Q leads to a decrease in V.

A capacitor is fully charged before the battery is disconnected. Then a dielectric material is placed between the plates of a capacitor. What happens to the capacitance (C), voltage (V) and charge (Q)?Single choice

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