∙In a rectangular coordinate system, a positive point charge q=6.00 nC is placed at the point x=

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

Key Concepts

Coulomb's Law

This law describes the electric force between two point charges, stating that the magnitude of the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. It provides the basis for calculating the electric field generated by a single point charge, which is essential for understanding electric fields in more complex distributions.

Electric Field of a Point Charge

The electric field produced by a point charge is defined as the force per unit charge experienced by a small positive test charge placed in the field. It is calculated using Coulomb’s law and is represented as a vector field, indicating both magnitude and direction, which points radially away from a positive charge and towards a negative charge.

Superposition Principle

This principle states that the net electric field created by multiple charges is the vector sum of the electric fields produced by each individual charge. It allows for the analysis of complex charge configurations by breaking the problem into simpler parts where the contributions from each charge are calculated separately and then summed.

Vector Decomposition

When dealing with electric fields in a coordinate system, it is necessary to break down the field into its x and y components. This process involves using trigonometric functions to resolve a vector into its horizontal and vertical components relative to the coordinate axes, which is crucial for accurately analyzing the field at specific points.

Symmetry in Electric Fields

Utilizing symmetry can simplify the calculation of electric fields in certain charge configurations. When charges are symmetrically placed, their individual fields may have components that cancel each other along certain directions. Recognizing these symmetries helps to reduce the complexity of vector addition and can make it easier to determine the net electric field.

Transcript

00:01 In this problem, we're going to be repeatedly applying this formula, which gives us the electric field, given the distances and the charge.

00:08 And we're also going to use the fact that the electric field points towards the minus charges and away from the positive charges.

00:15 And so let's start with part a.

00:19 In part a, we have the situation that we're at the origin here.

00:24 We want to find the electric field at the origin.

00:26 To our right, we have q1, and it's at a distance.

00:30 0 .15 meters.

00:33 And to our left, we have another charge, q2, also at a distance 0 .15 meters.

00:39 And in this case, they're both negative charges.

00:42 And so the electric fields will point towards them.

00:45 There would be an electric field here, e1, due to q1.

00:49 And electric field here, e2, due to q2.

00:52 Because of the symmetry with the distances and the charges, we know that e1 is going to be equal to e2.

01:00 But because they point in different directions, i'm going to add a minus sign here.

01:06 Moving the e2 over, i find that e1 plus e2 is equal to e total, which is equal to zero.

01:17 And so since they are in opposite directions, the resultant electric field is zero.

01:27 Let's draw the situation for part b.

01:29 So we still have q2 here, and we have q1 here, and it's 0 .15 meters.

01:39 Here, it's another 0 .15 meters here, and then it's another 0 .15 meters to our point, which i'll call p, that we're interested in finding the electric field at.

01:58 So now the electric fields are going to point away.

02:06 So that's e1 and that's e2.

02:09 So let's figure out how their magnitudes relate to one another.

02:14 So using the formula on the first page there, we have that e1.

02:18 Is equal to 2 ,397 nons per culum.

02:26 And that's when we use 0 .15 as the distance, since that's the distance from p to q1 here.

02:35 Now e2, we're going to use a larger distance.

02:38 We're going to use 3 times 0 .15, since that's the distance from p to q2.

02:43 And when we plug this in, we're going to get 266 newt mons per culum.

02:49 Notice that the larger distance between p and q2 is causing the electric field to be much less than e1.

02:57 To figure out the total electric field, e total in the x direction, we're going to sum these two up.

03:06 And when we do that, we get 2 ,600 newtons per cool.

03:12 And it's in the positive x direction where this is the positive extraction here.

03:18 And so that's the answer to part b.

03:21 Part c is where the problem gets a little tougher because we have to worry about the y component now.

03:27 So we have q2 and q1, let's say it's here.

03:33 Directly below q1 is the point that we're concerned with.

03:37 I'm going to call it point p.

03:40 And so there's going to be a y component here of e to y, which is the y component of the electric field due to this guy.

03:51 And there's also going to be an x component of the electric field due to this guy, e2x.

03:57 And then that covers this guy.

03:59 With q1, there is no x component because p is directly vertical to it, but there is a y component.

04:05 It.

04:06 And so there is an e1y here.

04:11 And in total, the electric field is going to point somewhere there.

04:17 That's e2 here, which is e2y and e2x combined.

04:22 And then we have to figure out what all three combined are.

04:26 And so to do that, let's figure out what e1 is.

04:32 E1, again, we just are worried about the vertical component here.

04:36 This distance, by the way, is 0 .4 meters.

04:41 And this distance here, between q2 and our point is 0 .5 meters.

04:48 And so e1 can be found just like before because it's only a vertical component, so we don't have to worry about the angles at all yet.

04:54 And we get that it's 337 newtons per cool.

05:03 And again, e1 x is equal to 0.

05:06 So e1y, well, run out of space here, is equal to the full part here, like that.

05:19 Now e2, which is this diagonal vector here, we can calculate like before.

05:26 And when we do that, we get 216 nons per kulum.

05:33 And now what we're going to do is calculate this angle here, the theta.

05:38 And then once we have the theta, then we can break this up into the horizontal and vertical components.

05:44 Now you can find the theta by using the sign function...

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