Answer :
Answer:
[tex]|F_3| = 5.4*10^{-6}N[/tex]
[tex]Direction 56.31^{0}[/tex] W from -ve x-axis
Explanation:
electrostatic force is given by
[tex]F =\frac{k*Q1*Q2}{R^2}[/tex]
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}
[tex]F_13 =\frac{k*q1*q3}{r_{13}^2} (-i)[/tex]
force between q2 and q3 is attractive and towards +ve y-axis, which is
[tex]F_23 = \frac{k*q2*q3}{r_{23}^2} j[/tex]
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
[tex]F_3 = F_{13} (-i) + F_{23} (j)[/tex]
F_3 =[tex]\frac{9*10^9*6*10^{-9}*5*10^{-9}}{0.30^2} (-i) +\frac{9*10^9*1*10^{-9}*5*10^-{9}}{0.10^2} j[/tex]
[tex]F3 = -3*10^{-6}i + 4.5*10^{-6}j N[/tex]
magnitude of force can be calculated as
[tex]|F3| = \sqrt{(- 3*10^{-6})^2 + (4.5*10^-6)^2}[/tex]
[tex]|F_3| = 5.4*10^{-6}N[/tex]
Direction [tex]= tan^{-1}\frac{Fy}{Fx} = tan^{-1}\frac{4.5}{3} = 56.31^{0}[/tex] W from -ve x-axis
Direction = 180 - 56.31 = 123.69 anti- Clockwise from +ve x-axis
The magnitude of the net electrostatic force on the third charge is calculated as follows is 5.41 x 10⁻⁶ N and the direction is 56.3⁰.
The given parameters;
- Charge Q1 = 6.0 nC
- Charge Q2 = -1.0 nC
- Charge Q3 = 5.0 nC
The electric force on charge Q3 due to charge Q1 is calculated as follows;
[tex]F_{13} = \frac{kq_1q_3}{r_{13}^2} (i)\\\\F_{13} = \frac{(9\times 10^9) (6\times 10^{-9})(5\times 10^{-9})}{(0.3)^2} \ (i) \\\\F_{13} = (3\times 10^{-6} \ i) \ N[/tex]
The electric force on charge Q3 due to charge Q2 is calculated as follows;
[tex]F_{23} = \frac{kq_2q_3}{r_{23}^2} \ (-j)\\\\F_{23} = \frac{(9\times 10^9)\times (-1\times 10^{-9})\times (5\times 10^{-9})}{(0.1)^2} \ (-j)\\\\F_{23} = (-4.5 \times 10^{-6} \ j)N[/tex]
The net electrostatic force is calculated as follows;
[tex]F_{net} = \sqrt{(3\times 10^{-6})^2 + (-4.5\times 10^{-6})^2} \\\\F_{net} = 5.41 \times 10^{-6} \ N[/tex]
The direction of the force is calculated as follows;
[tex]\theta = tan^{-1} (\frac{F_y}{F_x} )\\\\\theta = tan^{-1} (\frac{-4.5 \times 10^{-6}}{3\times 10^{-6}} )\\\\\theta = 56.3 \ ^0[/tex]
Thus, the magnitude of the net electrostatic force on the third charge is calculated as follows is 5.41 x 10⁻⁶ N and the direction is 56.3⁰.
Learn more here:https://brainly.com/question/15395424