Imagine three point charges at the corners of an isosceles...

Imagine three point charges at the corners of an isosceles triangle: qâ‚ = 2.22 x 10â»Â¹â° C, qâ‚‚ = 3.33 nC, and qâ‚ƒ = 4.44 x 10â»â¸ C. The charges qâ‚ and qâ‚‚ are 1.00 m apart and form the triangle's base. The triangle is 0.250 m tall. If qâ‚ƒ is at the top, what is the magnitude and direction of the resultant force on qâ‚ƒ? (4.08 x 10â»â¶ N, 150.2Â°)

Transcript

00:01 Alright, hello, in this question we're given this charge setup.

00:02 We have an equilateral triangle, which means that each of these angles is 60 degrees, and they have a side length of 20 centimeters.

00:09 We have three negative charges, and we're interested in the net electric field at this point c, which is halfway between charges 1 and 3.

00:19 So in order to figure out what this is going to be, i'm going to start by drawing the vectors that are each of the electric fields from these.

00:26 So i know that for a negative point charge, i'm going to have an electric field that is pointing towards it radially.

00:33 So for charge 1, which i'll draw here in black, i'm going to have an electric field, sorry, charge 3, not charge 1.

00:40 I'm going to have e3, which is pointing directly towards charge 3.

00:44 For charge 1, i'm going to have the same thing.

00:47 I'm going to have e1 is pointing directly towards it.

00:49 And then e2, this blue one, is going to be pointing directly towards that, and that's going to be e2.

00:56 Now if i want to just find the magnitude, i can set my axes to be whatever i want.

01:02 And so i'm going to make a very clever choice of axes here.

01:04 I'm going to say that that is the positive x direction, and then that is the positive y direction.

01:09 And because i know these are at right angles to one another, this is just going to make my life a lot easier, since i'm only asked for the magnitude.

01:17 So i know that my electric field in the x direction is going to be, well, i have e3 pointing in the positive direction, and i have e1 pointing in the negative direction, so that's going to be a negative.

01:29 What is the electric field generated by a point charge? well, it's going to be k times that point charge.

01:36 So for charge 3, it's going to be kq3 over the distance it is away squared.

01:41 So in this case, it's a distance a over 2 away, and that's going to be squared.

01:47 And then for e1, i have k times q1 over a over 2 squared as well.

01:53 Note here, i've accounted for the fact that these are negative charges already by just having the direction of my electric fields point towards those point charges.

02:04 So when i plug in q1 and q3, i don't need to plug in the sine.

02:11 I can go ahead and reduce this down, factor out some like terms, and i'll get q3 minus q1.

02:16 So that's going to be e in the x direction...

Question

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