Point charge A is on the x -axis at x=-3.00 cm. At x=1.00 cm on the x -axis its electric field

Point charge A is on the x -axis at x=-3.00 cm. At x=1.00...

Key Concepts

Effects of Charge Sign on Field Direction

The sign of a charge determines the direction of the electric field it creates. Fields from positive charges point away from the charge, while fields from negative charges point towards the charge. This understanding is crucial when determining the net direction of the electric field in systems with multiple charges.

Superposition Principle

The superposition principle in electrostatics states that the net electric field generated by a collection of point charges is the vector sum of the electric fields produced by each charge individually. This principle allows for the analysis of complex charge configurations by considering one charge at a time.

Vector Nature of Electric Fields

Electric fields are vector quantities, meaning that they have both magnitude and direction. When calculating the net electric field due to multiple charges, the direction of each individual field must be taken into account using vector addition.

Electric Field Due to a Point Charge

This concept explains that a point charge creates an electric field in the surrounding space which decreases in magnitude as the distance from the charge increases. The field's magnitude is given by Coulomb's law, showing an inverse-square dependence on the distance.

Coulomb's Law

Coulomb's law quantifies the electric force between two point charges. It states that the magnitude of the force (and thus the related electric field) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Transcript

0:00 Hello everyone.

00:02 This problem, we are asked to find the electric field at the origin in a one -dimensional setup where we are told that we have a charge qa that is sitting three centimeters away from the origin in the negative direction and a charge qb, which is sitting at five centimeters away from the origin in the positive direction.

00:24 And we are also told that the charge, the episode value of the charge of b is twice the absolute value of the charge of a.

00:35 And what we're also told apart from that is that the electric field of magnitude coming from a, from qa, at one centimeters to the right of the origin, is 2 ,700 units per couloms.

00:50 So then we're asked to find what is the electric field at the origin.

00:55 Given various scenarios for the signs of the charges of a and b.

01:00 So first what we have to do is we have to find the charge of a and b, and only then can we actually work out what the electric field will be at the origin.

01:10 So since we're told that the electric field at one centimeter away from the origin to the right, i .e.

01:16 Four centimeters away from charge a is 2 ,700 units per coulombs.

01:21 And we know gullum's cost is 8 .99 times into 9 utons, meter, square, for gullum squared, we can actually just work out what the electric field tells us in terms of what the charge of a is.

01:36 So we say that the electric field magnitude due to a point charge is k times the absolute value of the charge, divided by the separation between the point of interest and the charge squared, in which case that i've called that r .a.

01:50 So that's what that's our four centimeters.

01:53 And so we can rearrange that for qa and putting in the values, we find that qa is 4 .81 times 10 to minus 10 kulams.

02:03 So now that we have that, we can actually work at both charges.

02:06 So the charges that we're going to be looking at are qa equal to plus minus 4 .81 times 10 to 10 to 10 kulams.

02:16 And qb being twice that, so being plus minus 9 .62 times the number.

02:20 Minus 10 kulombs.

02:22 All right, so then we can use this information and put it back into our expression for the electric field.

02:28 Now remember that the electric field is a vector.

02:30 Right.

02:31 So we have to add these quantities, these electric fields coming from charges a and b as vectors.

02:37 So here i've written out what the electric field is going to be.

02:40 So it's going to be in brackets k so coolants constant times the charge in a divided by the coordinate of a squared minus k times the charge.

02:50 Charge of b divided by the coordinate of b squared and this is pointing in the x head direction.

02:58 Now you might wonder why i have this minus sign here, but so let's go back up and try to explain where that comes from.

03:05 So where does this come from? so this comes from the fact that if, you know, the positive sign is kind of easy to understand.

03:11 If i have a charge sitting at the origin and it's positive, then if a a has a charge that's a charge that's, is negative, then i'm going to get that this expression, k times qa, remember, q is less than zero, so i'm going to have an electric field that's pointing to the negative direction.

03:35 So there's going to be something, basically qa is going to be pulling the charge from the origin to itself towards the negative x's.

03:42 Now, similarly, if i have the same configuration with b being positive, then remember that the test charge that i put at the origin is positive.

03:52 Positive.

03:53 So i need the direction of the electric field to be pointing away from the origin still.

03:58 So it needs to be pointing towards the negative direction.

04:01 And so if i put a positive point charge at b, then i need that to be basically pushing the charge from the origin towards the negative part of the axis.

04:12 And that's where that minus sign comes from.

04:14 So now that that's clarified, we can basically just simplify this expression.

04:19 We can factorize k...

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