<p>A point charge q1=+2 &#956;C is located on the positive y axis at y=0.3m, and

Question

<p>A point charge q1=+2 &#956;C is located on the positive y axis at y=0.3m, and an identical charge q2 is located at the origin. Find the magnitude and direction (angle with respect to the x axis) of the total force that these two charges exert on a third charge q3=+4 &#956;C that is on the positive x axis at x=.4.</p>

Explanation / Answer

ok so the force from 1 on 3 is: ok so we have: k = 9e9 q1 = 2 q2 = 2 q3 = 4 distance between q1 and q3 = (.4^2+.3^2) = .5 m distance between q2 and q3 = .4 m F (1 on 3) = k q1*q3/l^2 = 9e9*2*4/(.5)^2 = 2.88e11 N in some diagonal direction F (2 on 3) = k q2*q3/l^2 = 9e9*2*4/(.4)^2 = 4.5e11 N in the direction straight right so im guessing at this point we need to convert the first force into its components. so we need the angle theta which will be tan^-1(.3/4) = 36.86 degrees. so now that we have that we can break up our force into two components. 2.88e11*cos(36.9) = x component of F(1>3)= 2.304e11 2.88e11*sin(36.9) = y component of F(1>3) = 1.728e11 ok now that thats settled we combine the horizontal forces so 2.304e11+4.5e11 = total x component of F on 3 which is 6.804e11 and then total y component is just the remainder which is 1.728e11. so we multiply trig style. a^2+b^2 = c^2 and we get the final force of 7.0180355e11 or simply = 7.018e11 N

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