A +5.0 µC charge is placed at the origin and a -3.9 µC charge is placed at x = 25 cm. At what coordinates can a third charge be placed so that it experiences no net force?
E1=k•q1/(x+ Δx)² (to the right)
E2= k•q2/(Δx)² (to the left)
E1= E2
k•q1/(x+ Δx)² = k•q2/(Δx)²
q1•(Δx)² = q2• (x+ Δx)²= q2• (x²+ 2•x•Δx + Δx²)
5• Δx²=3.9(0.0625+0.5 Δx + Δx²)
1.1 Δx² -1.95• Δx -0.24375 =0
Δx = {1.95±√(1.95²+4•1.1•0.24375)}/2.2=
=(1.95±2.21)/2.2= 1.89 m and -0.12m (extraneous root)
Δx =1.89 m
To find the coordinates where a third charge can be placed so that it experiences no net force, we need to calculate the net force on the third charge due to the other two charges.
The net force on a charge due to electric forces is given by Coulomb's Law:
F = k * |q1 * q2| / r^2
where F is the force, k is the Coulomb's constant (k = 9.0 x 10^9 N*m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
Let's consider the two charges:
Charge 1 (q1) has a value of +5.0 µC (5.0 x 10^-6 C) and is placed at the origin (x = 0).
Charge 2 (q2) has a value of -3.9 µC (-3.9 x 10^-6 C) and is placed at x = 25 cm = 0.25 m.
We want to find the coordinates (x, y) where a third charge can be placed to experience no net force.
Since we are given the x-coordinates of the two charges, we can assume the third charge lies on the x-axis as well. So, the y-coordinate of the third charge will be 0.
Let the charge at the origin (q1) be charge A, and the charge at x = 25 cm (q2) be charge B.
The net force on the third charge (charge C) due to A and B can be calculated separately, and these forces should cancel each other out for the net force to be zero.
F_net = F_AC + F_BC
F_AC = k * |q_A * q_C| / r_AC^2
F_BC = k * |q_B * q_C| / r_BC^2
Here, r_AC and r_BC are the distances between charges A and C and between charges B and C, respectively.
Since both charges A and B are placed on the x-axis, we can directly calculate the distances as follows:
r_AC = x_C - 0 (distance of C from A on x-axis) = x_C
r_BC = x_C - 0.25 (distance of C from B on x-axis) = x_C - 0.25
Now, to find the coordinates (x, y) where a third charge can be placed to experience no net force, we equate the magnitudes of the two forces and solve for x:
F_AC = F_BC
k * |q_A * q_C| / r_AC^2 = k * |q_B * q_C| / r_BC^2
|q_A * q_C| / r_AC^2 = |q_B * q_C| / r_BC^2
|q_A * q_C| * r_BC^2 = |q_B * q_C| * r_AC^2
|q_A| * r_BC^2 = |q_B| * r_AC^2
|q_A| * (x_C - 0.25)^2 = |q_B| * x_C^2
Substituting the values:
|5.0 x 10^-6| * (x_C - 0.25)^2 = |-3.9 x 10^-6| * x_C^2
We can solve this quadratic equation to find the value(s) of x_C.
After finding the value(s) of x_C, we can use these coordinates to locate the third charge on the x-axis (since y-coordinate is 0) to experience no net force.