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A charge of  +.4 mC is at (-3, 0) meters and a charge of  +.9 mC is at...

A charge of  +.4 mC is at (-3, 0) meters and a charge of  +.9 mC is at (+1, 0) meters.  

What is the direction of the force felt by a +.5 mC charge placed at (0, 3) meters due to the original two charges?

Group of answer choices

a) 97.2 degrees below the (-x) axis

b) 97.2 degrees above the x axis

c) 75.6 degrees below the x axis

d) 97.2 degrees above the (-x) axis

e) 75.6 degrees above the x axis

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Homework Answers

Answer #1

Here we have given that,

Q1 = +0.4 mC is at (-3, 0) meters

and Q2 = +0.9 mC is at (+1, 0) meters

d1 = (3²+3²)^0.5 = 4.24264 m

Theta 1 = arctan(3/-3) = 45°

d2 = (3²+1¹) = 3.16227 m

Theta2 = 71.565°

So that for the direction of the force felt by a +0.5 mC charge placed at (0, 3) meters due to the original two charges will be given as,

Here the force due to q1 will be given as,

Fq1 = kq1q3/4.24264² × ( Cos45°i + Sin45°j)

Fq1 = (70.732)i + (70.732)j N

Now Fq2 = Kq2q3/d2² * ( -Cos71.565°i + Sin71.565j) N

Fq2 = -(128.07889)i + 384.2356jN

Here the total force on q3 will be,

Fnet = Fq2 + Fq2

And the direction of force felt by the q3 will be

Theta = arctan (Fnety/Fnetx) = 97.2° above the x axis

Hence b) 97.2 degrees above the x axis

Will be the correct answer.

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