Fundamentals of Physics Chapter 23 Solutions: Gauss’ Law
Halliday Resnick Volume 2 Solutions for Chapter 23 ‘Gauss’ Law’ is one of the most reliable resources you can use for strengthening Physics concepts about Gauss’ law and its applications. By reading Resnick Halliday Gauss Law, you will be able to understand about gaussian surfaces, about electric flux. You will also learn about how to find the electric field inside and outside the spherical Gaussian surfaces, in a uniformly charged sphere, in infinite nonconducting sheets, and so on. We suggest you go through our Halliday Resnick Physics Solutions for Class 12 once, all your concepts will be crystal clear.
Resnick Halliday Fundamentals of Physics Volume 2 Gauss’ Law consists of 55 numerical and word problems that let you understand the concepts associated with Gauss’ law. The questions are divided into 6 different modules. You will find questions on electric flux, charged-isolated conductor, also some questions related to Gauss’ law and its applications. Apart from this, some additional problems are also there in this chapter. All of these questions are covered beautifully, and solving such questions will not only aid you in scoring well in your Class 12 Board examination but also plays a significant role for you, ace the various toughest entrance examinations like JEE and NEET.
Halliday Resnick Fundamentals of Physics Volume 2 Solutions for JEE is an amazing book for you. Unlike other books, this book follows a different approach and contains a variety of questions, some are easy and some are quite tough at which you need proper guidance to tackle it. Therefore, we strongly suggest you read and solve all the problems of Halliday Resnick Physics Volume 2 Gauss’ Law with the help of our solutions. The best thing in our solutions is that it is prepared by the highly qualified and experienced subject experts, aiming that how easy and detailed solutions are provided to you so that you could understand the concepts easily.
Important Topics of Resnick Halliday Class 12 Volume 2 Solutions Chapter 23: Gauss’ Law
Well, In this Chapter 23 Gauss’ Law you will study the different concepts like the electric flux of an electric field, you will apply gauss law, using it you will determine the electric field generated by an infinite uniformly charged, insulating plane and insulated rod, also in a uniformly charged spherical shell. Further, you will apply Gauss’ law to find an electric field inside and outside charged conductors.
Now, let’s have a look at some of the important concepts and definitions associated with this chapter-
- Solid Angle-
You know, the Plane angle is the angle between two intersecting lines but if we talk about the solid angles. It is a different angle, It is an angle which is not formed between two intersecting lines, it is basically a 3-Dimensional angle that is formed between an object in a plane or 3-d space. The plane angles are measured in degrees or radians. However, a solid angle is measured in steradians.
- Electric Flux –
We know the nature of electric lines of force very well that it starts from positive charge and terminates into negative charge. In fact, these lines never intersect with each other. So here we can conclude the Electric flux, it is nothing, is the rate of the electric field in that given area. It can also be said that electric flux is directly proportional to the number of these electric field lines going through the virtual surface.
Let’s have a look at how electric flux varies in different conditions. Like, what is the nature of electric flux at the uniform electric field and in the non-uniform electric field how it behaves.
- For Uniform Electric Field–
In a uniform electric field, electric flux is passing through the surface(S) of vector area
Φ = E . S = E S Cos θ
Where E is the magnitude of the electric field, θ is the angle between the electric field lines and normal to S
- For Non-Uniform electric field-
For non-uniform electric fields, we take the component of surface area dS. Thus, the electric flux in this condition is the multiplication of the electric field to the component of the surface area.
d Φ = E . dS
Where E is the electric field and dS is the differential area.
Gauss’ Law describes the electric flux as the surface integral of the electric field and the component (dS) of the closed surface S.
Φ = ∬s E . dS
Where dS is the differential area of the closed surface it is very small.
- S-I Units –
Electric Flux has its units Volt-metre and in terms of dimensions, it is N m2 C-1 .
- Gauss’ law and its Applications
Before going to the applications of Gauss’ law just have a brief overview of Gauss’ law.
So, what is Gauss’ law?
A Gauss’ Law describes the relationship between the charges and electric fields. If Qe is the net charge in the closed Gaussian surface. Then according to the Gauss law, the net electric flux is directly proportional to the total charge enclosed in that closed surface.
In other words, Gauss law can also be expressed in a vector calculus in an integral form.
It is equal to the Φ = ∬A E . dA
Where E is the magnitude of the electric field and dA is the differential area of the closed Gaussian surface.’
- Using Gauss’ law you will determine the intensity of the electric field inside a uniformly charged sphere.
- You will also apply the gauss law for finding the intensity of the electric field at any point due to the infinite line of charge.
- Also, you will determine the electric field at the uniformly charged spherical shell with the radius R under three conditions when (R>= r), (R=r), and when (R<= r).
- Using Gauss’ law, you will compute the electric field at an infinite non-conducting sheet with the uniform surface charge density σ.
Exercises Discussion of Halliday Resnick Class 12 Volume 2 Solutions Chapter 23: Gauss’ Law
Module 1: Electric Flux
Module-23.1 has 3 questions based on the concepts of electric flux where you need to determine the electric flux through the plane surface and cubical gaussian surface.
Module 2: Gauss’ Law
Module-23.2 of Halliday Resnick Volume 2 Solutions Chapter 23 Gauss’ Law has 13 odd questions. You need to go through the gauss’ law and its applications to tackle the questions involved in this module.
Module 3: A Charged Isolated Conductor
In Module-23.3 of Halliday Resnick Physics Solutions, only 5 questions are there but you need to have a decent knowledge about a charged isolated-conductor. Questions use the concepts of it and can be tricky if you have a little knowledge in it.
Module 4: Applying Gauss’ Law: Cylindrical Symmetry
While in the questions of Module-23.4, you need to apply the gauss’ law in Cylindrical Symmetry. This Module has 11 questions based on the electric field at a drum surface, linear charge density, the magnitude of an electric field in the inner and outer shell, and electric field along two axes.
Module 5: Applying Gauss’ Law: Planar Symmetry
With 11 questions, this Module-23.5 will test you on how you apply the gauss’ law in planar symmetry. This includes questions on finding electric fields at different points, writing vector notation of electric field on a point and ratio of charges on a sheet.
Question no.36 in a Module-23.5 is about the two large parallel non-conducting sheets with the uniform charge density σ. You just need to determine the electric field above, between and below the sheets.
Module 6: Applying Gauss’ Law: Spherical Symmetry
The last Module-23.6 of the Halliday Resnick Volume 2 Solutions for Physics contains 12 such questions based on the application of gauss law in Spherical Symmetry. Questions are based on finding charge on a sphere and charge in the inner and outer shell.
Apart from this 25 additional problems are also in this chapter that has a mix of questions almost from all the concepts of the chapter.
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