Two parallel-plate capacitors, 4.9 μF each, are connected in parallel to a 70 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors

Answer: a) 343 μC. b) same as a)

Explanation:

If the capacitors are connected in parallel, this means that both have the same voltage between the plates, i. e., 70 V from the battery.

By definition, the capacitance of each capacitor can be written as follows:

C₁ = Q₁ / V and C₂ = Q₂ / V

Now, if the capacitors and in parallel, and if C₁ = C₂, ⇒Q₁ = Q₂ = C₁V = C₂V

Replacing by the values for C₁, C₂ and V, we have the following:

Q₁ = Q₂ = 4.9. 10⁻⁶ F. 70 V = 343 μC.

So, the total charge transferred by the battery to the capacitors is the sum of the charges on both capacitors:

Q = Q₁ + Q₂ = 2. 343 μC = 686 μC.

Now, let's suppose that C₁ is squeezed in such a way that the separation between plates is halved.

For a parallel-plate capacitor, we can write an expression for the capacitance, as follows:

C = ε A / d

where ε = dielectric constant of the insulating material between plates.

A = Area of one of the plates

d = separation between the plates.

We can see that d is halved, C will double.

If C is doubled, and V remains constant, Q must increase in the same proportion, i. e. must be double.

As the other capacitor remains without any changes, it will not receive any additional charge.

So the additional charge delivered by the battery will be equal to the original charge (because it will go double), i. e., 343 μC.

Now, due to the charge conservation principle, as charge can't be created or destroyed, and can't accumulate neither, the increase in the total charge stored on the capacitors, must be equal to the additional charge transferred by the battery to C₁ once it changed his value after squeezing it.