A point charge Q is placed at the origin. A second charge, 2Q, is placed on the

Question

A point charge Q is placed at the origin. A second charge, 2Q, is placed on the x axis at x = –3.0 m. If Q = 50 µC, what is the magnitude of the electrostatic force on a third point charge, –Q, placed on the y axis at y = +4.0 m?

I need help figuring out how to work this out.

This is the worked out solution that my professor posted for the problem, but I don't fully understand how he did most of this, or how he got the numbers in the third and fourth parts. Any help in understanding this better would be apprieciated.

Explanation / Answer

q1 = 50 mciro C (at origin)

q2 = 100 micro C (at x = -3 m)

q3 = -50 micro C (at y + 4m)

F31 = k*q3*q1/d13^2

= 9*10^9*50*10^-6*50*10^-6/4^2

= 1.40625 N (downward)

F31x = 0

F31y = -1.40625 N

F32 = k*q2*q1/d12^2

= 9*10^9*100*10^-6*50*10^-6/(3^2 + 4^2)

= 1.8 N (towards q2)

angle made by F21 with -x axis, theta = tan^-1(4/3) = 53.`13 degrees

F32x = -1.8*cos(53.13) = -1.08 N

F32y = -1.8*sin(53.13) = -1.43 N

so,

F3x = F31x + F32x = 0 - 1.08 = -1.08 N

F3y = F31y + F32y = -1.40625 - 1.43 = -2.83625 N

so, F3 = sqrt(F3x^2 + F3y^2)

= sqrt(1.08^2 + 2.83625^2)

= 3.03 N <<<<<<<<<<-------------------Answer

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