Two charged particles are located on the x axis. The first...

Key Concepts

Vector Addition in One Dimension

When dealing with electric fields in one dimension, such as along a single axis, it is crucial to account for both magnitude and direction. In this context, vector addition reduces to algebraic addition of scalar quantities with signs, which allows for a simpler analysis of how different fields combine to produce a net field at a point.

Charge Sign and Field Direction

The sign of a charge determines the direction of its electric field: fields radiate outward from positive charges and converge inward toward negative charges. This property is important when performing vector addition of fields, as the direction (positive or negative along a chosen axis) affects whether the fields reinforce or oppose one another.

Coulomb's Law

Coulomb’s law describes how the electric field created by a point charge depends directly on the magnitude of the charge and inversely on the square of the distance from the charge. This principle is fundamental in understanding how individual charges contribute to the field at a point in space, and it provides the mathematical basis for calculating the field produced by any isolated charge.

Electric Field Superposition

The principle of superposition states that the net electric field produced by multiple charges is the vector sum of the individual electric fields contributed by each charge. This concept allows for the analysis of complex charge distributions by considering the individual effects separately and then combining them to determine the overall field.

Transcript

00:01 Okay, so this is the graph i drew for this question.

00:03 So positive q is located at the left of the origin points at the distance a.

00:09 And uncharged q2 is placed at the right of the origin points at the distance 3a.

00:14 And e1 is the electric field due to the positive charge and the direction of e1 has point away from the positive charge, which is point toward the right.

00:22 And we know the net electric field had the magnitude of 2kq over a square.

00:27 So therefore, the possible value, for q2 should be 2 because it's either positive values or negative values.

00:40 If it's positive, that means the electric field due to the q2 charge will be moving, will be directly away from the q2.

00:50 If it's negative charge, then the electric fuel due to the q2 charge will be directly toward the q2.

00:57 So let's think about the first possibility, which is if q2 is positive.

01:10 Well, so if it's positive, we have e nets is equal to e1 minus e2.

01:21 So e1 is the electric view from a positive charge q.

01:24 E2 is the electric field from the unknown charge q2.

01:28 The reason why i use minus here is because the direction of the electric view from unknown charge is opposite with the direction of positive of positive, okay, since both of them are positive in this case.

01:39 So therefore, we have e1 minus e2 is equal to 2 kq over a square.

01:56 And then when e1 is equal to kq because it's a positive q, okay? and the distance between the positive charge and the origin point is a.

02:16 So just a square.

02:19 And this thing minus e2 is equal to 2kq over a square.

02:32 Remember e2 is equal to k kx constant k times charge k2 over 3a square okay because the distance between k2 and the origin point is 3a so therefore we can plug in back to the equation here now we have kq over a square minus kk2 over 3a to the power 2 and a equal to 2 kk 2 q a square and this will give us negative kq2 over 9a square is equal to kq over a square because i move the kq over a square from the left side to the right side so therefore i would have kk over a square on the right side and as you can tell we can cancel k and a square in this case therefore we have negative q2 divided by 9 is equal to q.

03:47 So therefore we have q2 is equal to negative 9q.

03:55 So this is the first possible values.

03:58 Now let's take a look at the second possibility, which is when if q2 is negative...

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