I am asking this question twice (see separate questions) - answer it correctly t

Question

I am asking this question twice (see separate questions) - answer it correctly twice and I will rate life saver 2x!!!!

This is part of Example 15.5 from the Serway Vuille Book.

(a) Place a charge of -5.80 µC at point P and find the magnitude and direction of the electric field at the location of q2 due to q1 = 7.35 µC and the charge at P.
q2 = -5.00 micro C.

.
magnitude 5 N/C
direction 6°

(b) Find the magnitude and direction of the force on q2.
magnitude 7 N
direction 8°



Please provide explanation + correct final answers, thank you!!

Explanation / Answer

The electric field from a point particle is found by using E = kq/r2, however, there are two different charges, q1 and the one at point p, and each must be solved separately.
Let find the E from q1 first.

E(q1) = ( 9e9 )( 7.35e-6 C ) / ( .3 m )2

E(q1) = 735,000 N/C. The direction is TO THE RIGHT ( positive q1 has E fields that point AWAY from q1 )

Now, lets find E due to charge at p.

E(p) = ( 9e9 )( 5.8-6 C ) / ( .5 m )2

E(p) = 208800 N/C. The direction is TOWARDS point p, since the charge is negative, and negative fields point towards the charge. This means the E(p) is at an angle, we need to find the angle. We know enough about the legs of the triangle, so we use:

tan = .4 m / .3 m

= 53.13 degrees.

Now, we need to take E(p) and break it into its x and y components.

208800 N/C cos(53.13) = 125280.3 This is the x piece of E(p), and it points to the LEFT

208800 N/C sin(53.13) = 167039.8 This is the y piece of E(p) and it points UP.

Now we add all the x fields, of which there are two, 125280.3 LEFT and 735,000 RIGHT = 609719.7 RIGHT

Now we put the total x and total y back together by using pythagorean theorem:

609719.72 + 167039.82 = c2

c = 632,187 N/C

The angle of this final E field can be found using

tan = 167,039.8 / 609,719.7

= 15.32 degrees above the poisitive x direction.

B) Now that we know the E, the force is easier to find. We use F = qE, where q is the charge we will PLACE on the E we just solved for. We are placing q2, which is -5e-6 C, so:

F = ( 5e-6 C )( 632,187 N/C )

F = 3.161 N

The direction of the force is OPPOSITE the directon of the E field ( r neg charges always go AGAINST E field directions ). Since the E field is pointing 15.32 above the positive x axis, then q2 is going 15.32 below the neg x axis, which can be rewritten as 195.32 degrees.

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