please! help! Three point charges are located at the corners of an equilateral t

Question

please! help!

Three point charges are located at the corners of an equilateral triangle as shown. Each leg of the triangle has a length of 1.38 meters. And the charges have values as follows. Q1 = 4.61 microcoulombs, Q2 = - 8.69 microcoulombs, and Q3 = -6.09 microcoulombs.

Calculate the net electric force (both magnitude and direction) exerted on Q1.

Magnitude?  Direction?

Calculate the net electric force (both magnitude and direction) exerted on Q2.

Calculate the net electric force (both magnitude and direction) exerted on Q3.

Magnitude?  Direction?

Explanation / Answer

F12 = k*q1*q2/L^2

= 9*10^9*4.61*10^-6*8.69*10^-6/1.38^2

= 0.189 N

F13 = k*q1*q2/L^2

= 9*10^9*4.61*10^-6*6.09*10^-6/1.38^2

= 0.133 N

F23 = k*q1*q2/L^2

= 9*10^9*8.69*10^-6*6.09*10^-6/1.38^2

= 0.25 N

a) Net force on Q1, Fnet1 = sqrt( F12^2 + F13^2 + 2*F12*F13*cos(60))

= sqrt(0.189^2 + 0.133^2 + 2*0.189*0.133*cos(60))

= 0.28 N (towards -y axis)

b) Net force on Q2, Fnet2 = sqrt( F12^2 + F32^2 + 2*F12*F32*cos(120))

= sqrt(0.189^2 + 0.25^2 + 2*0.189*0.25*cos(120))

= 0.226 N (60 degrees above -x axis)

c) Net force on Q3, Fnet3 = sqrt( F13^2 + F32^2 + 2*F13*F32*cos(120))

= sqrt(0.133^2 + 0.25^2 + 2*0.133*0.25*cos(120))

= 0.217 N (60 degrees above +x axis)

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