An air capacitor is made by using two flat plates, each with...

An air capacitor is made by using two flat plates, each with area $A,$ separated by a distance $d$ . Then a a metal slab having thickness a (less than $d$ ) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.66). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance $C_{0}$ when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits $a \rightarrow 0$ and $a \rightarrow d$ .

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a gap. Its capacitance is directly proportional to the area of the plates and inversely proportional to the separation between them, with the simple formula C = ??A/d for an air-filled capacitor. This fundamental concept is key to understanding how changes in geometry affect the storage of electrical energy.

Series Combination of Capacitors

When a conductor such as a metal slab is inserted between the plates without making contact, the capacitor is effectively divided into multiple regions that act as separate capacitors connected in series. The overall capacitance is determined by the reciprocal of the sum of the reciprocals of the individual capacitances. This method of combining capacitors is crucial for analyzing complex capacitor arrangements.

Effect of Conducting Insertion

Introducing a conducting slab within a capacitor alters the effective distance over which the electric field is established. Although the slab itself does not store energy like the dielectric gaps, it changes the potential distribution by creating new boundaries. Understanding how inserting a metal slab changes the effective geometry is essential for designing and modifying capacitor structures.

Limit Analysis in Capacitance

Investigating the limits as the thickness of the inserted metal approaches zero or the full separation distance provides insight into the continuity and behavior of the capacitance formula. In the limit of a negligible slab thickness, the capacitor behaves almost like the original capacitor with no insertion, while in the opposite limit, the effective gap for energy storage diminishes, significantly altering the overall capacitance. This analysis helps in verifying the correctness of derived formulas and understanding extreme cases in capacitor design.

Transcript

00:02 Hi, so we are given a problem of parallel platy capacitor.

00:06 So we have an air capacitor which is made by two metal plates, each having area a and are separated by distance t.

00:15 So let me draw the metal plates here.

00:19 So we have two parallel metal plates, each having area of air and these two pender plates are separated by distance t.

00:33 Now it's given that in between the two metal plates, a metal slab of thickness a which is less than d is inserted so we have a metal slab of thickness a this thickness is less than the separation d and it is inserted in between the two plates and in the first part of the problem we need to find the net capacitance of this arrangement now note that in between the two parallel plates so in between these two plates we have a metal slab insert it so this is a metal slab and not a dialectic slab that is why this arrangement is like an arrangement of two capacitors connected in series so we have a capacitor consisting of the upper so let me write it like this so we have a capacitor consisting of the upper plate and the metal slab so let's call this capacitor 7 and then we have another capacitor consisting of the middle metal slab and the lower plate so this is capacitors c2 which is filled with so this two capacitors are filled with eo and the distance as you can see between the upper plate and the metaslav and the metaslav and the lower plate let's see this let's call this distance bt prime and for both the capacitors the distance is same and if you will see the distance d prime is going to be equal to total distance which is d minus the distance where we have the metal slab.

02:33 So the thickness of the metal slab which is a divided by 2.

02:38 So this is the thickness or the distance between the two capacitors, t prime.

02:46 Now having found that, so in the first part of the problem, we need to find the equivalent capacitance of this arrangement of the capacitors which is in which in series so the equivalent capacitance can be written as one over c because since it's like a series combination so 1 over c c equals 1 over c1 plus 1 over c2 now note that c1 and c2 they are they are separated by the same distance so and they have the same area so c1 is indeed equals to c2 which is equal to x0 a divided by the distance which we have already calculated here d prime which is d minus a or divided by or divided by 2 so that's c1 and c2 now since c1 is equal to c2 which is going to imply the capacitance c is nothing but it's going to be equal to half times c1.

04:10 So the equivalent capacitance is half of c1 and we have already calculated c1 which is equal to c2.

04:18 So from this, we find that the equivalent capacitance for this combination is going to be epsilon not a divided by d minus a...

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