Question

An empty parallel plate capacitor is connected between the terminals of a 5.12-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

Answer 1

Answer:

V₂ = 2* V₁ = 10.24 V

Explanation:

• By definition, the capacitance is given by the following relationship between the charge on one of the plates (for a parallel plate capacitor) and the voltage between them:

       $C = \frac{Q}{V} (1)$

• Applying (1) to the capacitor, once fully charged and disconnected from the battery,  V = V₁ = 5.12 V.
• Now, we know that for a parallel plate capacitor, the capacitance is independent from the voltage applied, as follows:

       $C = \frac{\epsilon_{0} * A}{d} (2)$

• where ε₀ = 8.85*10⁻¹² F/m, A is the area of one of the plates, and d is the spacing between plates.

• If we double the spacing between the plates d, the capacitance will be reduced to half, due to d₂ = 2* d₁
• Once disconnected from the battery, due to the principle of conservation of the charge, the charge Q must remain constant, i.e.,

       Q₁ = Q₂ = Q

• Since the only variable that can be modified is the voltage V, if the capacitance reduces to the half of the original value, the voltage must be doubled in order to keep  C₂ = C₁/2, true.

       ⇒ Q/V₂ = Q/V₁ * 1/2 ⇒ V₂ = 2* V₁ = 2* 5.12 V = 10.24 V