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:
- 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:
- 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.