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# Fundamentals of Physics Chapter 22 Solutions: Electric Fields

Halliday Resnick and Walker Volume 2 Solutions for Chapter 22 ‘Electric Fields’ is one of the most recommended books for all JEE aspirants. The salient points discussed in Resnick Halliday Physics solutions Electric Fields are the basics of an electric field; its definition and formula, what are electric field lines, the definition of an electric dipole, and its formula. After going through this chapter you will be able to explain the origination and termination of electric field lines, draw a charged particle and find its direction and determine whether it is a positively or negatively charged particle, draw and identify a dipole axis, and many such concepts around electric fields.

Resnick Halliday & Walker Fundamentals of Physics Volume 2 Solutions for Electric Fields has a total of 101 questions divided into different 3 sets of exercises. While solving these questions of Electric Fields, you would be applying various formulas for how to calculate field due to a point charge, how to find field due to continuous charge distribution and due to a charged disk. You would also get to know how to measure the force on a point charge and a dipole in an electric field.

Resnick Halliday and Walker Fundamentals of Physics book for Class 12 is something you would enjoy reading due to its lucid language and beautiful explanations of each and every concept. Solutions for Resnick Halliday & Walker Electric Fields provided by our academic experts would help in clearing these concepts thoroughly with to-the-point solutions. If you are facing any issue in clearing the complicated topics then taking the help of our solutions would surely help you cross those hurdles and prepare you better for any exam.

## Important Topics for Halliday Resnick and Walker Volume 2 Solutions Chapter 22 Electric Fields

• Electric Field – Any electric charge sets up a field around its space which is its electric field. This field exerts a force on other objects around it and also experiences the force set up by other charges in its location.
• Source Charge – This is the charged particle which is generating the electric field.
• Test Charge – This is the object that is experiencing the force generated by the source object.

If a positive test charge ch+ is placed then the electric field on that charge is measure by: = /ch+Here E is the electric field, F is the electrostatic force.

• Electric Field Lines – An electric field is a vector point function since it has a definite magnitude and direction at every field point. Electric field lines are a way to visualize the direction and magnitude of the electric field. These lines start at positive charges and terminate at negative charges.
• Electric Field Vector – This is tangent to the electric field line at any point in the electric field. If the magnitude of the electric field is high at a point then the number of vector lines crossing that point would also be high.
• Field Because of a Point Charge – A point charge Ch sets up an electric field at a distance d from it, given by:

E (electric field) = 1/4πε0 (|Ch| / d2)

Here Ch = Charge which is generating the electric field

d = distance from the point charge

ε0 = permittivity of space = 8.854 X 10-12 C2 N-1 m-2

• Field Because of an Electric Dipole – A pair of equal and opposite charges, say +Ch1 and –Ch1 separated by a distance d, form an electric dipole. It’s an electric dipole moment ( ) is given by Ch1d and points from the negative charge to the positive charge. The dipole axis runs through both the charges and the magnitude of the electric field set up at a distant point from this is given by:

E = (1/2πε0)( /d13 ), where is the dipole moment and d1 is the distance of the point from the centre of the dipole.

• Field Because of Continuous Charge Distribution – For an object like a rod, where there is an infinite number of charged particles uniformly distributed, one can find the continuous charge distribution of these particles by finding the electric field due to differential elements of the rod and then using calculus integration to find the net vector.

• Field Due to a Charged Disk – A uniformly charged disk with a radius r and a surface charged density of σ exerts an electric force along the central axis of the disk and its magnitude is given by:

E = σ/2ε0(1 – (d1/d12 + r2)), Here d1 is the distance of the point along the axis from the centre of the disk

• Torque on a Dipole within an Electric Field – The behaviour of a dipole in an electric field can be described by the 2 vectors; electric field vector and dipole moment vector, without knowing anything about the dipole’s structure.
• Torque on the dipole • The potential energy of dipole due to its position in the field:
• ### Discussion of Exercises of Resnick Halliday and Walker Volume 2 Solutions Chapter 22: Electric Fields

Questions

The first part of the exercise has a total of 14 problems. These questions need you to calculate electric field magnitude and direction, the linear momentum of charges in the vicinity of other charged particles, ranking arrangements based on the magnitude of the electric field, potential energies of a dipole, ranking of electrically charged disks based on their electric field around a point, and the force due to continuous charge distribution.

Module 1: The Electric Field

The first module has 2 questions where you need to apply the formula of an electric field to solve them.

Module 2: The Electric Field Due to A Charged Particle

The 2nd module of Electric Fields in Resnick Halliday and Walker fundamentals of physics volume 2 solutions has 15 questions. You get to solve problems on uniform charge distribution, unit vector notation due to 2 charged particles along an axis, the electric field around a charged disk. For some problems, you also need to draw a sketch of the electric field lines around the particles.

Module 3: The Electric Field Due to a Dipole

The 3rd module has 4 questions all based on the electric dipole. In these questions you need to apply binomial expansion on the equation of the electric field of a dipole, calculate the magnitude and direction of the dipole’s electric field on a point along an axis. There is also a question on electric quadrupole which is made of 2 electric dipoles.

Module 4: The Electric Field Due to a Line of Charge

The 4th module has 12 questions where you get to apply knowledge on uniform charge distribution to solve these problems. The charge density over objects like a disk, sphere, and ring is given and you need to find volume charge density.

Module 5: The Electric Field Due to a Charged Disk

The 5th module has 5 questions on the concept of finding the magnitude of the electric field on circular surfaces.

Module 6: A Point Charge in an Electric Field

The 6th module of Electric Fields Chapter has 17 questions. The formula for electric force on a point charge within an electric field has to be used to solve these problems. These questions vary from medium to high complexity so that you get full exposure to the kind of question which can come on this topic.

Module 7: A Dipole in an Electric Field

The 7th module has 6 questions on calculating torque experienced by a dipole within an electric field.