Point charge q1 500 nc is at the origin and point charge q2...

Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00nC is on the x -axis at x = 3.00cm. Point P is on the y-axis at y = 4.00cm. Part A Calculate the electric fields E⃗ 1 and E⃗ 2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors (see example 21.6 in the textbook). Express your answer in terms of the unit vectors i^, j^. Enter your answers separated by a comma. Part B Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form. Express your answer in terms of the unit vectors i^, j^.

Transcript

00:01 Hello, here we have to solve the following problem.

00:02 Point charge q1, that is negative 5 nanoculums, is located at the origin and the point charge q2, positive 3 nanoculums, is on the x -axis at x2, which is, i should add x, that is 3 centimeters.

00:25 Meanwhile, point p is on the y -axis at 4 centimeters.

00:33 Let's solve part a.

00:35 In part a we have to calculate the electric field, e1 and e2, due to the charges at point p.

00:46 Let's do this.

00:48 Let's sketch x's y and x, and here we will show q1 at the origin and at point p.

00:59 Which is at 4 at y this q1 creates the field which is directed up oh sorry that's actually yeah let's be more accurate with the direction q1 is negative that's why e1 is directed downwards meanwhile point q point 2 is a tax of 3 and that's positive that's why it creates the field which is directed from this q2.

01:38 So here we have to look at this triangle formed by y and x.

01:46 And the diagonal of this triangle d is square root of x squared plus y squared.

01:52 So let's calculate this d.

01:54 That is a square root of 9 plus 16 centimeters wishes 5 centimeters.

02:15 Let's now label alpha.

02:19 And as we can notice sign of alpha is y over d that is 0 .800 or g or z yeah 0 .800 and cosine of alpha is 3 over 5 that is 0 .600 so now we know everything to calculate the field so but first here we have to calculate e1 let's start with e1 because that's the simplest.

03:00 E1, let's first calculate its magnitude.

03:04 That is kq1 over y squared.

03:28 And that is divided by y squared, which is 4 centimeters squared, or 0 .0 400 meters squared.

03:42 Let's complete this calculation.

03:59 That is 28 ,000 125 newton percolum.

04:09 And now we have to express it in terms of the unit vectors.

04:13 So then e1 is negative 28 ,000 125 newton per coulomb times vector i, which is rough, which is roughly negative negative.

04:29 2 .81 times 10 power by 4 newton proculump times vector i guess so that is done now we have to calculate e2 has two components let's first calculate e2 x x x is negative kq2 over d squared times cosine alpha then e to x is k constant over 5 centimeters which is squared multiplied by the church few two and multiplied by cosine of alpha let's calculate it that is negative 600, 6 ,880 newton per coulump.

06:16 And now let's calculate e2y.

06:19 E to y is positive and that is kq2 over d squared, sine alpha.

06:58 This is 302 ,400 ,000 400 newton per coulomb...

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