Charge q1 = 6.0 nc is at (0.30 m, 0), charge q2 = -1.0 nc is at (0, 0.10 m), and charge . what are the magnitude and direction of the net electrostatic force on the 5.0-nc charge due to the other charges?

W from -ve x-axis

Explanation:

electrostatic force is given by

if both charge have different signs, Force will be attractive, and if both charge have same signs,  force will be repulsive .

Now Net force will be

force between q1 and q3 is repulsive and towards -ve x-axis, which is

where q3 = 5.0nC {GIVEN}

force between q2 and q3 is attractive and towards +ve y-axis, which is

Now given that

q_1 = 6 nC, q_2 = -1 nC, q_3 = 5.0 nC

r_13 = 0.30 m

r_23 = 0.10 m

total force will be

F_3 =

magnitude of force can be calculated as

Direction W from -ve x-axis

Direction = 180 - 56.31 = 123.69 anti- Clockwise from +ve x-axis

answer is c

explanation:

c) up

hope this