# 3 charged particles are placed at the corners of an equilateral triangle of side d = 1.15 m. Q1 = +4.0 µC Q2= -6.0 µC Q3 = -6.0 µC. Calculate the...

3 charged particles are placed at the corners of an equilateral triangle of side *d* = 1.15 m. *Q*_{1} = +4.0 µC *Q*_{2}= -6.0 µC *Q*_{3} = -6.0 µC.

Calculate the magnitude and direction of the net force on each due to the other two.

Force on Q_{1}: |
N at | ° counterclockwise from +x axis (to the right) |

Force on Q_{2}: |
N at | ° counterclockwise from +x axis (to the right) |

Force on Q_{3}: |
N at | ° counterclockwise from +x axis (to the right) |

### 1 Answer | Add Yours

The charge on Q1 by Q2 and Q3 are (1/4pie0)*Q1Q2/d^2 and (4pie0)^(-1)*Q1Q3/d^2 = 0.163327N

9*10^9*4*6*10^(-12)/1.15^2 = 0.163327N attraction as the charges are different, and

9*10^9*4*6*(10^(-12)/1.15^2 = 0.163327N attraction as the charges are different.

Therefore the net force is of the two equal magnitude force with 60 deg between by law of parallelogram is (0.163327)2cos30 = 0.282890717 N with 270 degree to X axis.

In a similar way we can show that the force on Q2 is 0.282990717N acting 30 degree to X axis.

In a similar way we can show that the force on Q3 is 0.282990717N acting 150 degree to X axis.