Question

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab (dielectric constant 2.0) is then inserted between the plates, and completely fills the gap.
(a) What is the change in the charge on the positive plate when the Teflon is inserted?

Answer 1

Answer:

The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.

Explanation:

• It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:

       $C = \frac{\epsilon*A}{d}$

• Where ε, is the dielectric constant of the material that fills the space between plates.
• When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
• At the same time, the capacitance of a capacitor, by definition, is as follows:

       $C =\frac{Q}{V}$

• If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
• As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
• So, the change in the charge of the positive plate is +2.5 nC.