A parallel plate capacitor is charged and the charging battery is subsequently disconnected. lf the plates of…

Analyze the question: A charged capacitor is disconnected from the battery, so the charge has nowhere to go. Then the plates are pulled farther apart. We must decide what happens to the capacitance and the charge.

Concept: For a parallel plate capacitor:

C = ε₀A / d

• C = capacitance
• A = plate area
• d = separation between the plates
• ε₀ = permittivity of free space

Capacitance is inversely proportional to the plate separation.

Solution:

1. The battery is disconnected, so the charge Q stays constant. There is no path for charge to flow on or off the plates. This rules out options C and D.
2. The plates are moved farther apart, so d increases.
3. Since C = ε₀A/d and A and ε₀ do not change, increasing d makes C decrease.

Extra insight: with Q constant and C decreasing, the potential difference V = Q/C increases. The energy stored, ½QV, also increases. This is because you do work pulling the plates apart against their attraction.

Checking the options:

• A. Capacitance would increase: wrong, since d increases.
• B. Capacitance would decrease: correct.
• C. Charge would increase: wrong, since the battery is disconnected.
• D. Charge would decrease: wrong, since the charge is trapped on the isolated plates.

**Answer: B.

Exam tip: Ask yourself whether the battery is connected or disconnected.

• Disconnected: Q stays constant.
• Connected: V stays constant.

Also remember that C ∝ 1/d, so moving plates apart always lowers capacitance.