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.
The term used is " WORK".
I'm really sorry it's not one of the choices.
For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.
So, for Q5, we can first find out the voltage across R₂ by V=IR.
Voltage across R₂ = 2.5 × 8 = 20V
Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.
So, by V=IR,
current of R₃ = 
= 5A
Q6. voltage across R₁ = 2 × 4 = 8V
∴voltage across R₂ = 8V
current of R₂ = 
= 1A
<h3><u>Alternative method</u></h3>
From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.
ie. for Q5,
= 
= 
I₃ = 5A
Q6. 
= 
= 
I₂ = 1A
<span>(c) energy travels from the object at higher temperature
to the object at lower temperature.
Size and mass have no effect.</span>
