Dielectrics in Capacitors, Equivalent Capacitance Explained – Part 2

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In Part-1: Learning about Capacitors, we already got an idea about the functioning of the capacitors. But now, with that understanding, how to make use of them in electric circuits?

This is what we focus on in this article. Getting inclined towards the numerical aspect of the capacitors is our goal for this article.

1. Clearing the basics

As discussed in part-1, the more the charge separation happens (+Q on one plate and -Q on the other), the more will be the potential difference across the plates of the capacitor(V). So basically we can say, 

$Q \propto V$

Now we introduce the ‘constant’ of proportionality –> Capacitance (C). The relation becomes :

$C = \frac{Q}{V}$

Note that :

• Increasing the charge doesn’t increase the capacitance of the capacitor as it may appear in the above equation. For simplicity, remember it as –> Once a capacitor is made, its capacitance value is stamped on it !!
• Instead, capacitance is decided by the physical dimensions and the dielectric used. (For understanding, once the vessel is manufactured, it has some dimensions to it, and based on those dimensions, you can decide the capacity of the vessel.)

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2. Parallel Plate Capacitors

There are many configurations possible to make a capacitor, but the simplest one to analyze is the Parallel plate Capacitor. This setup basically consists of :

• 2 metal plates of area ‘A’ kept at a distance ‘d’ apart from each other
•  Dielectric medium (of dielectric constant ‘k’) inserted between the plates
• Important Condition : (A>>d)

The capacitance of the parallel plate capacitor is given by :

We can clearly verify from the above expression that the capacitance just depends upon the geometrical factors and the dielectric constant.

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3. What is a dielectric, and what is its use?

“In simple words, dielectrics are a type of insulating materials which allow Electric field but doesn’t allow an electric current to pass through them.”

They are mainly used to serve 3 purposes: 

Purpose of using Dielectric (Expand This)

There are 3 purposes to use a dielectric –

1. A dielectric acts as an insulating material that maintains the gap between the plates
2. It increases the maximum voltage that can be applied across the capacitor plates without getting breakdown
3. It increases the capacitance of a capacitor

3.1 Maintaining Gap

It is necessary to maintain the gap (even though small) between the metal plates. Because the capacitor would lose all of its storing capacity if the metal plates come in contact with each other, since in that case, it would just behave as a simple conductor.

3.2 Increasing the maximum voltage without breakdown

Dielectric Breakdown: 

• Each dielectric material has its own breakdown voltage. 
• If the applied voltage becomes greater than the breakdown voltage of the dielectric material, the atoms start to get ionized, and we know that ions do conduct electricity. 
• Because of this, the whole capacitor starts to act as a conductor 

The better the dielectric material, the higher the dielectric breakdown voltage. Let’s consider 2 situations :

The dielectric breakdown voltage of the material (say D2) is more than that of air. This implies that more voltage across the capacitor plates is required in the case of D2 for the breakdown to happen. This proves our point that the insertion of a dielectric allows us to apply more voltage across the capacitor plates without causing any dielectric breakdown.

3.3 Increasing Capacitance

Suppose we have a capacitor with just air between the plates. Now we insert a dielectric material of dielectric constant ‘k’ between the plates completely. Have at the look at the flowchart below to just get a quick overview of what happens!

4. Series Combination in Capacitors

“Like the way we have current in case of resistors, in the same way, for capacitors, we have charge.”

• For Capacitors to be said in a series combination, the charge flowing through them should be the same.

Consider 3 capacitors C₁, C₂, and C₃ in series combination, and V₁, V₂, and V₃ are the potential differences across them, respectively.

Since we need to find the ‘equivalent’ capacitance :

$V_{\text{eq}} = V_{1} + V_{2} + V_{3}$

$\frac{q}{C_{\text{eq}}} = \frac{q}{C_{1}} + \frac{q}{C_{2}} + \frac{q}{C_{3}}$

But the charge flowing through all is still the same. Hence, we get:

$\frac{1}{C_{\text{eq}}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

So, we can observe that, by keeping capacitors in series, we get the value of equivalent or resultant capacitance, which is even less than the one that has the least capacitance among the three. Suppose,  C₂ < C₁ < C₃, then C_eq < C₂ 

5. Parallel Combination in Capacitors

Again:  “Like the way we have current in case of resistors, in the same way, for capacitors, we have charge.”

• For the Capacitors to be in Parallel, the potential difference across all should be the same.

Consider 3 Capacitors C1, C2, and C3 in parallel combination, and the charges passing through them are q1, q2, and q3, respectively.

We know the relation from KCL : 

$q = q_{1} + q_{2} + q_{3}$

$C_{\text{eq}}(V) = C_{1}(V) + C_{2}(V) + C_{3}(V)$

But the potential difference across all is still the same. So, we get:

$C_{\text{eq}} = C_{1} + C_{2} + C_{3}$

The equivalent capacitance incase of a parallel combination will be greater than the greatest among the three (here)

6. Methods to simplify the circuits

Now, there are again some types of network circuits in which we are expected to find the equivalent Capacitance. 

There are several methods to simplify and solve such type of circuits, like :

• Mirror Symmetry
• Folding symmetry
• Voltage method (Rearrangement)

We have already looked at the above methods with context to resistors in Part-2: Combining Resistors.

Though the article is for Resistors, the approach of simplifying the circuit/network still remains the same!

How to approach the problem?

1. Simplify the given complex circuit using the methods discussed already in the linked article, i.e., Mirror Symmetry, Folding Symmetry, etc.
2. You will get the simplified circuit in the form of series and parallel connections
3. Use the discussed concept to solve the series and parallel combination of capacitors

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7. Special equivalent capacitance problem :

Example: Find the equivalent capacitance across points A and B

Step-1 :

Distribute the voltages across all plates. In this example, we consider the voltage/potential at A to be ‘a’ and at B to be ‘b’. Still, we are not able to cover all the plates. So we introduce another unknown potential ‘x.’

Remember that: Potential always remains constant on a conductor

Step-2 :

Assign the numbers to each face of the plates 

Step-3 :

Keep points ‘a’ and ‘b’ at the ends, and all the unknowns which we introduced should come in between.

Step-4 :

To make a capacitor, we need 2 plates and separate them by a distance

• Faces 2 and 3 make a capacitor
• Faces 4 and 5 make a capacitor
• Faces 6 and 7 make a capacitor
• Faces 8 and 9 make a capacitor

So, now, we just look at the numbers assigned to their faces and make a simplified circuit.

Step-5 :

Solve by normal Series- Parallel concepts

The equivalent capacitance of the simplified circuit is 5C/3,  where C is :

$C = \frac{A \,\varepsilon_{0}}{d}$

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Conclusion:

• So, with this, we are done studying about combining the given capacitances in various patterns to get the required capacitance in our circuit. Also, in previous articles, we have learnt about Resistors – Part 1: Dealing with Resistors and Part 2: Combining Resistors
• Just combining these 2 components – Resistors and Capacitors opens up a whole new set of things that can be developed. We will be looking into these in our upcoming articles.

Till then, Keep Learning!