Two positive charges, each 4.18 mu*C and a negative charge, -6.36 , are fixed at the vertices of an equilateral triangle. Calculate the length of the triangle side if the net electrostatic force on the negative charge is F net =27.4 N. Give your answer units

Consider the point charges shown below. What will be the magnitude and direction of the force on the 1.0µC charge if the -2.0µC and -3.0µC charges are held in place? -2.0 με 16.0 cm (+ 12.0 cm 1.0 μC -3.0 µC Part A The magnitude of the force on the 1.0uC will be: nνα ΑΣφ F = N Submit Request Answer Part B The point charge will experience the force at an angle of: Πνα ΑΣφ ?

Two point charges Q1 and Q2 are 2.10 m apart, and their total charge is 14.0 µC. If the force of repulsion between them is 0.0982 N, what are magnitudes of the two charges? Enter the smaller charge in the first box. Q1 6.10 uc Q2 = 7.90 uC You are correct. Your receipt no. is 161-6663 Previous Tries If one charge attracts the other with a force of 0.0709N, what are the magnitudes of the two charges if their total charge is also 14.0 µC? The charges are at a distance of 2.10 m apart. Note that you may need to solve a quadratic equation to reach your answer. Enter the charge with a smaller magnitude in the first box. Q1 %3D Q2 Submit Answer Incorrect. Tries 7/10 Previous Tries

A charge of -2.70 nC is placed at the origin of an xy-coordinate system, and a charge of 2.20 nC is placed on the y axis at y = 3.70 cm. Part B Find the magnitude of this force. F = Part C Find the direction of this force. 0 = I Part A If a third charge, of 5.00 nC, is now placed at the point x = 2.70 cm, y = 3.70 cm find the x and y components of the total force exerted on this charge by the other two charges. Express answers numerically separated by a comma. Fa, Fy= N N ° below the +x axis