A negative point charge q1 = -4.00 nC is on the x-axis at x = 0.60 m. A second point charge q2 i

step 1 So let's first write down the known quantities in the problem. So we have a point charge Q1 that

A negative point charge q1 = -4.00 nC is on the x-axis at x...

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that quantifies the electric force between two point charges. It states that the magnitude of the force is proportional to the product of the charges and inversely proportional to the square of the distance separating them. This law is used to derive the electric field produced by a point charge, which is essential in solving problems involving electric fields.

Electric Field from a Point Charge

The electric field created by a point charge is determined using Coulomb's Law, expressed as E = k|q|/r², where E is the electric field magnitude, k is the electrostatic constant, q is the charge, and r is the distance from the charge. The field's direction depends on the sign of the charge: it points away from positive charges and toward negative charges.

Superposition Principle

The superposition principle in electromagnetism states that the net electric field at any point is the vector sum of the electric fields created by each individual charge. This principle is crucial when evaluating the combined effects of multiple point charges, as it allows one to calculate the overall electric field by adding the contributions from each charge while taking into account both magnitude and direction.

Influence of Distance on Electric Field

The magnitude of the electric field from a point charge diminishes with the square of the distance, according to the inverse square law. This relationship means that even small changes in distance can significantly affect the magnitude of the field, an important consideration when calculating net fields from charges placed at different locations.

Transcript

00:01 So let's first write down the known quantities in the problem.

00:05 So we have a point charge q1 that is equal to negative 4 manaculum.

00:13 And this is placed on the x -axis at a distance 0 .6 meter from the origin.

00:21 And i'm calling that x -1 so that it corresponds to the distance of q1.

00:27 And then we have another charge q2 whose magnitude is.

00:32 Unknown and we need to find that and this q2 is placed at a distance let's say x2 that is given to be 1 .2 meter and this is also from the origin of the x -axis and we need to find the sign and magnitude of q2 for so that the net electric field is 50 muten per coulum in the positive x -axis, the net electric field at the origin due to both q1 and q2 is 50 newton per coulum towards the positive x -axis.

01:24 And in the second case, the net electric field at the origin should be 50 newton per column but in the negative x -axis.

01:35 So in both the cases, the magnitude of the electric field is same but the directions are different.

01:43 So, for the first case, let's find e1 and e2, that is the electric field due to each of these charges.

01:57 So the net electric field at the origin, since e net is at the origin, so we need to find e1 and e2 at the origin only.

02:07 So e1 will be equal to k times q1 over x1 square because this is the distance from the origin.

02:18 Substituting the values with k to be equal to so k is the column constant and this is equal to 9 times 10 to the power 9 newton times meter square over column square so now substituting the values over here we get e1 to be 100 newton per column now similarly we can find e2 and one thing we know for sure is the the second charge, the electric field due to second charge must be in the negative x direction because so that the second charge is negative and its magnitude has to be 50 newton per column.

03:09 So, e1, e2 is actually k times q2 over r2 square.

03:23 So one thing i forgot to mention is the direction of e1.

03:26 So let's mark the vectors so that we can include the direction over here.

03:35 So let me draw the axis.

03:38 So it will be easier for you guys to understand.

03:41 So here is the point where we are finding the elective field and q1 is somewhere around here.

03:48 Q2 is somewhere around here.

03:50 And both of them are negative.

03:52 Not both.

03:53 Q1 is negative.

03:54 So the electrical field at p should be towards q1 because q1 is negative.

04:02 So this is in the positive x direction.

04:08 So this is actually i -cap as a direction i -cap.

04:15 And e2, let's say that q2 is positive for now.

04:25 So the electric field due to e2 will be in the negative x direction because it will be away from q2 since q2 is positive so negative x and we don't know the value of e2 yet but we do know the net electric field so we know that the net electric field is actually 50 newton per column in the positive x -axis so i cap and by superposition principle we say that the net electric field is actually the vector sum of e1 and e2.

05:06 So this means that 50 newton per coulum icap is actually equal to 100 newton per column i cap plus kq2 over r2 square negative i cap.

05:26 So this is for e2.

05:28 Let me move on to the next page.

05:31 So from here we can say that e2 vector which is kq2 over r2 square r2 or x2.

05:43 Okay, so this is actually i have taken this to be x, not r.

05:51 So let me just write x here.

05:56 So over here as well, the direction.

06:09 So i'm basically solving this equation...