In a rectangular coordinate system a positive point charge q=6.00 ×10^-9 C is placed at the poin

step 1 So the direction of the two fields are shown here. So this is E1, and this is E2, and the dot is

In a rectangular coordinate system a positive point charge...

In a rectangular coordinate system a positive point charge $q=6.00 \times 10^{-9} \mathrm{C}$ is placed at the point $x=+0.150 \mathrm{~m}, y=0,$ and an identical point charge is placed at $x=-0.150 \mathrm{~m}, y=0 .$ Find the $x$ - and $y$ -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) $x=0.300 \mathrm{~m}, y=0$ (d) $x=0, y=0.200 \mathrm{~m}$ (c) $x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m} ;$

In a rectangular coordinate system a positive point charge $q=6.00 \times 10^{-9} \mathrm{C}$ is placed at the point $x=+0.150 \mathrm{~m}, y=0,$ and an identical point charge is placed at $x=-0.150 \mathrm{~m}, y=0 .$ Find the $x$ - and $y$ -components, the magnitude, and the direction of the electric field at the following points: (a) the origin;
(b) $x=0.300 \mathrm{~m}, y=0$
(d) $x=0, y=0.200 \mathrm{~m}$
(c) $x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m} ;$

Key Concepts

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the magnitude of the electric force between the charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is fundamental for calculating the electric field generated by a point charge and understanding interactions in electrostatics.

Electric Field of a Point Charge

The electric field due to a point charge is a vector quantity that represents the force per unit charge experienced by a test charge placed in the field. Its magnitude can be calculated using the equation E = k|q|/r^2, where k is Coulomb's constant, q is the charge, and r is the distance from the charge to the point of interest. The field points radially outward for a positive charge and radially inward for a negative charge.

Superposition Principle

The superposition principle in electrostatics states that the net electric field created by multiple charges is the vector sum of the electric fields produced by each charge individually. This principle is crucial when analyzing systems with more than one point charge, as it allows one to compute the total field by adding the contributions of each charge at any given point in space.

Vector Components

Breaking down an electric field into its vector components (typically along the x and y axes) is essential for solving problems in a rectangular coordinate system. This technique involves using trigonometric functions to resolve the electric field into horizontal and vertical components, which can then be combined to determine the overall magnitude and direction of the field.

Coordinate Systems in Physics

A coordinate system, such as the rectangular (Cartesian) coordinate system, provides a framework for specifying the position of charges and points in space. It facilitates the analysis of problems involving vectors by helping to define directions and determine distances between points. In problems involving electric fields, the coordinate system is used to describe the location of charges and the points where the fields are evaluated.

Transcript

00:02 So the direction of the two fields are shown here.

00:10 So this is e1, and this is e2, and the dot is a.

00:16 So e1 is equal to e2, which is equal to 1 over times the absolute value of q over r squared, r equals 0 .150 meters.

00:34 So e equals e2 minus e1.

00:39 Which equals 0, and ex equals 0, and ey equals 0.

00:50 For part b, we have, the two fields have the directions shown in this figure.

01:02 So this is e2, and then this is b, and this is e1.

01:06 So e is equal to e1 plus e2 in the same.

01:13 Sorry, in the plus x direction.

01:25 So e1 is equal to 1 over 4 pi epsilon times that's the value of q1 over 4 over r squared.

01:41 And that is 8 .988 times 10 to the 9 times 6 times 10 to the negative 9.

01:51 Over 0 .15 squared and that leaves you with 2396 .8 in divided by c.

02:05 E2 is 1 over 4 pi epsilon not times the absolute value of q2 divided by r squared and that is 8 .988 times 10 to the 9th times 6 times 10 to the 9th divided by 0 .45 squared.

02:28 That leaves you with 266 .3 in per c.

02:33 So e equals e1 plus e2, which is 2396 .96 .8 plus 26 .3.

02:45 That leaves you with 2660 ends per c, and x is equal to plus 2660 in divided by c, and ey is equal to 0.

03:05 So for c, we have sine of theta is equal to 0 .4 meters divided by 0 .5 meters, that is 0 .8...

Please give Ace some feedback