Two 2.00-μC point charges are located on the x axis. One is at x=1.00 m, and the other is at x=-

Two 2.00-μC point charges are located on the x axis. One is...

Key Concepts

Vector Addition

Vector addition is a mathematical method used to combine multiple vectors such as electric fields or forces. Because electric fields are vectors, both their magnitude and direction must be considered when summing contributions from different sources.

Point Charges

Point charges are idealized charged objects whose size is negligible compared to the distances between them. This simplification allows the use of Coulomb's law directly without having to consider the geometry or spatial distribution of the charges.

Superposition Principle

The superposition principle states that the net electric field or force experienced at a point due to multiple charges is the vector sum of the fields or forces created by each individual charge. This principle is essential when calculating fields or forces from more than one source.

Electric Force

Electric force is the interaction between charged objects and is given by Coulomb's law. It is a vector quantity describing both the magnitude and direction of the force that one charge exerts on another. It can be attractive or repulsive depending on the signs of the charges involved.

Electric Field

The electric field is a vector field that represents the force per unit charge that a positive test charge would experience at any point in space due to other electric charges. It is calculated by dividing the force acting on the test charge by the magnitude of the charge and is measured in newtons per coulomb (N/C).

Coulomb's Law

Coulomb's law quantifies the electric force between two point charges. It states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, with the constant of proportionality being the Coulomb constant.

Transcript

00:01 So in this question, we are told that we've got two microculum point charges that are located on the x -axis.

00:09 That's what i've shown here in red.

00:12 And then we're asked to determine the electric field on the y -axis at y equals 0 .5 meters, which is the point that i've shown here in blue.

00:21 And then if we place a charge there, at that point, what is the force on that charge? so let's begin by calculating the electric field at this point p, which is located 0 .5 meters away from the origin at the y axis.

00:38 So first of all, what i'd like to point out is the symmetry here.

00:42 So if you look at the electric field coming from the left -hand charge, that's going to be directed up and towards the right.

00:51 And then if you look at the electric field that is coming from the right -hand charge, that's going to be directed.

00:57 Pointed up and towards the left and because these points are you know equal distance from point p and the charges are the same magnitude the electric field in the x are in the x direction is going to cancel out here so the x components of electric field one will will cancel with the x component of electric field two so ex is going to equal zero, so we don't need to worry about that.

01:32 We only need to calculate the electric field in the y direction.

01:44 So what we're going to be doing is we're going to be taking a component of the electric field in the y direction.

01:53 And let's label this angle here as theta, just to help us out with that.

01:59 And i'm going to draw another triangle now that looks something like this.

02:06 And the reason why i want to draw that triangle is because then we can label the distance between the charge and the point p.

02:16 And i can also point out that the angle there that's picking out the y direction is actually going to be the same as this angle here in this triangle.

02:28 So that's going to be quite useful for our calculation in a second.

02:33 Okay, so let's calculate the electric field in the y direction.

02:42 So we know that everything is going to be doubled here because the electric field from the point on the left -hand side is going to be the same as the electric field on the point from the point on the right -hand side.

02:54 So i'm just going to add a factor of two in here.

02:57 And then the electric field from each point charge will be kq over r squared.

03:04 And again, we want to pick out the y direction.

03:08 So we're going to multiply everything here by cost theta.

03:12 And if you look at the larger red triangles, you can see that costata is actually equal to y over r.

03:21 So i'm going to replace that cost theta with 2kq.

03:27 Sorry, we're going to replace the cos theta with y over r...