L=19 cm, d=3.8 cm, x=0.8cm. Please show all work I rate right answers thanks Mak

Question

L=19 cm, d=3.8 cm, x=0.8cm. Please show all work I rate right answers thanks

Make sure to show all your work and explain clearly what you are doing. This problem is due at the beginning of your exam on Friday. You can earn a maximum of 2 percentage points for it. I will be picky. Get values for L, A, and k from the tables below. Variable Capacitance: A variable capacitor can be constructed by inserting a slab of dielectric material partway into the space between a pair of charged parallel plates. One such capacitor has plates of length L and area A, and the plates are separated by a distance d=2.00mm. If the dielectric constant of the inserted material is k, then what is the capacitance as a function of the distance x that the dielectric material is inserted into the space between the plates?

Explanation / Answer

Hi

Data

It is useful to know that the capacitance of a parallel plate capacitor is:

C = eo A/d ; where A is the area of the plate, d is the distance between them and eo is the vacuum permittivity, which value is 8.85*10-12 C2/Nm2.

Another important fact is that when you have capacitors in parallel, the equivalent capacitance is the sum of the capacitances of the capacitors:

C = C1 + C2.

Finally, the effect of a dielectric inside a capacitor is to increase the value of the capacitance in certain proportion. Said proportion is dictated by the value of k:

C = kCo ; where C is the value of the capacitance, Co is the value of the capacitance without the dielectric and k is a property of the material used as dielectric, known as dielectric constant (its value is always bigger than 1).

Procedure

The trick with this kind of problem is to imagine the original capacitor as a combination of two capacitors. In this case we can imagine two capacitors in parallel, one with a dielectric and the other without it. If we do that, we have the following:

C = C' + Co ; where C is the total capacitance of the capacitor, C' is the capacitance of the part with dielectric and Co is the capacitance of the part without dielectric.

As one introduces the capacitor the value of x increases, and the value of L- x goes down. The area of the parallel plates, should change according to the value of lenght, so it must be a function of that parameter.

For instance, if the parallel plates are squares, then the area should be:

A(x) = x2 :::::: so the total area should be ::::::: A(L) = L2 ::::::: and the area of each part of the capacitor should be: A(x) = x2 (for the part with dielectric) and A(L-x) = (L-x)2 (for the part without dielectric).

If the area is a rectangle of measures L*b, then the function should be:

A(x) = x*b ; so:

A(L) = L*b ; total area :::::::: A(x) = x*b ; capacitor with dielectric :::::: A(L-x) = (L-x)*b ; capacitor without dielectric.

The list goes on and the only thing that change is the way to calculate A, but it is clear that in any case this value can be turn into a function of the lenght, and said function has interesting properties:

A(x) = f(x) :::::::: where f(0) = 0 and f(L) = Ao ; where Ao is the total area of the capacitor.

Once that is clear, the function of the capacitance in terms of x is easy to figure:

C(x) = (eo/d)*[ A(L-x) + kA(x) ] ; where 0 < x < L

To provide a numerical answer you should know the function A(x), which depends on the shape of the parallel plates (a square, a rectangle, a circle, or something more complex). if the shape is too complex, it might be necessary to integrate.

Therefore, knowing L, d and x (in a certain instant) is not enough to give a numerical answer.

I hope it helps.

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