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
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.
Get Answers For Free
Most questions answered within 1 hours.