a. Explanation Given: The electric field is, E = − ∇ φ Where φ is potentialfunction. The potential function is, φ = k Q r where r 2 = x 2 + y 2 + z 2 is the square of the distance between a variable point P ( x , y , z ) and k is constant. Calculation: The electric fieldis E = − ∇ φ (1) Let the function be φ = k Q r . Compute the value of r x as follows. r = x 2 + y 2 + z 2 r x = 1 2 x 2 + y 2 + z 2 ( 2 x ) = x x 2 + y 2 + z 2 Obtain the value of φ x as follows. φ = k Q r φ x = k Q ( − 1 r 2 r x ) = k Q ( − 1 r 2 ) ( x r ) = − k Q x r 3 Thus, the value of φ x is − k Q x r 3 . Compute the value of r y as follows. r = x 2 + y 2 + z 2 r y = 1 2 x 2 + y 2 + z 2 ( 2 y ) = y x 2 + y 2 + z 2 Obtain the value of φ y as follows. φ = k Q r φ y = k Q ( − 1 r 2 r y ) = k Q ( − 1 r 2 ) ( y r ) = − k Q y r 3 Thus, the value of φ y is − k Q y r 3 . Compute the value of r z as follows. r = x 2 + y 2 + z 2 r z = 1 2 x 2 + y 2 + z 2 ( 2 z ) = z x 2 + y 2 + z 2 Obtain the value of φ z as follows (b) To determine To show: The electricfield at a point has a magnitude | E | = k Q r 2 and explain this relationship.

2nd Edition

ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 12.6, Problem 77E

To determine

To show: The three dimensional electric field due to a point charge is given by E(x,y,z)=kQ〈xr3,yr3,zr3〉 .

(b)

To determine

To show: The electricfield at a point has a magnitude |E|=kQr2 and explain this relationship.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.6, Problem 76E

Chapter 12.6, Problem 78E

Students have asked these similar questions

Software Test Technology

None

Phyton The surface of the Earth is curved, and the distance between degrees of longitude varies with latitude. As a result, finding the distance between two points on the surface of the Earth is more complicated than simply using the Pythagorean theorem. Let (t1, g1) and (t2, g2) be the latitude and longitude of two points on the Earth’s surface. The distance between these points, following the surface of the Earth, in kilometers is: distance = 6371.01 × arccos(sin(t1) × sin(t2) + cos(t1) × cos(t2) × cos(g1 − g2)) The value 6371.01 in the previous equation wasn’t selected at random. It is the average radius of the Earth in kilometers. Create a program that allows the user to enter the latitude and longitude of two points on the Earth in degrees. Your program should display the distance between the points, following the surface of the earth, in kilometers. Hint: Python’s trigonometric functions operate in radians. As a result, you will need to convert the user’s input from degrees to…

Chapter 12 Solutions

CODE/CALC ET 3-HOLE

Ch. 12.1 - Give two pieces of information which, taken...Ch. 12.1 - Find a vector normal to the plane 2x 3y + 4z =...Ch. 12.1 - Where does the plane 2x 3y + 4z = 12 intersect...Ch. 12.1 - Give an equation of the plane with a normal vector...Ch. 12.1 - To which coordinate axes are the following...Ch. 12.1 - Describe the graph of x = z2 in 3.Ch. 12.1 - What is a trace of a surface?Ch. 12.1 - What is the name of the surface defined by the...Ch. 12.1 - What is the name of the surface defined by the...Ch. 12.1 - What is the name of the surface defined by the...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

The three dimensional electric field due to a point charge is given by E ( x , y , z ) = k Q 〈 x r 3 , y r 3 , z r 3 〉 .

Solution Summary: The author explains that the three-dimensional electric field due to a point charge is given by E=-nabla phi .

Videos

To show: The three dimensional electric field due to a point charge is given by E(x,y,z)=kQ〈xr3,yr3,zr3〉 .

To show: The electricfield at a point has a magnitude |E|=kQr2 and explain this relationship.

Want to see the full answer?

Chapter 12 Solutions