Capacitor Problems

What would happen if you keep a block of wood between the two plates of a capacitor? How would the capacitance associated with the system adjust to this ‘intrusion’? Read on to find out how the capacitance involving combinations of such ‘dielectrics’ is evaluated.

Based on the changes in the quantities observed, we classify that wood behaves as a dielectric of a certain dielectric constant 'k'. Even water or air or wool could replace wood in our example, each exhibiting a different ‘k’.

Now look at an example with three different dielectrics placed between the capacitor plates as shown. Let their dielectric constants be k₁,k₂ and k₃. Throughout the analysis consider the area of the capacitor plate shown below to be ‘A’ and the width of each capacitor,‘d’.

Our objective is to find the equivalent capacitance. Thinking in the language of circuits, what might be strikingly evident in this configuration is that the part with dielectric k₁ can be assumed to be in ‘series’ with the ‘parallel’ combination of dielectrics k₂ and k₃.

Using the formula for capacitance as:

we get,

and

The factor of half is contributed by the truncated area. The resultant of C₂ and C₃ in parallel, say C_{4 ,} is :

Now, this resultant is in series with the capacitor of capacitance C₁ of dielectric k₁.

Therefore the resultant capacitance, C, is :

which gives us,

There exists another way of looking at this.

Divide the dielectric k₁, along the line dividing k₂ and k₃. Now we visualize the system as a combination of four different capacitors. The way in which they are connected can be analyzed from the figure below.

Therefore the resultant capacitance can be found out by considering the series combination of k₁ and k₂ to be in parallel with the series combination of k₁ and k₃.

The capacitance C₅ of the series combination of first and second capacitors is,

where

Similarly C₆ can be determined_. Therefore, the net capacitance is,

After solving this problem in two different ways, we’ve realized that all roads have actually not led to Rome. Which one led us to the destination safely?

The second one. Let us see why.

In the first pathway we have considered that k₂ and k₃ are parallel. This is the flaw as the potential difference between the boundary dividing k₁ and k_2-k₃ need not be constant, i.e., it is not given to be an equipotential surface. By making them parallel we have violated this rule!

Note: As a special case if k₂ = k₃, both methods will yield correct answer. Convince yourself and find out why this anomaly rises! (Hint: Now, what happens to the potential throughout the boundary?)