Particle A of charge 3.00 x 10-4 C is at the origin, particle B of charge -6.00 x 10-4 C is at (4.00 m, 0)
and particle C of charge 1.00 x 10-4 C is at (0, 3.00 m). Find the magnitude and direction of the
resultant electric force acting on C.
X force on C = + k ( 6 )(1)(10^-8) (4/5)
note 4/5 is cos angle and unlike charges attract)
Y force on C = +k(3)(3)(10^-8) - k(6)(1)(10^-8)(3/5)
magnitude = sqrt (X^2+Y^2)
cos theta = X / Y
To find the magnitude and direction of the resultant electric force acting on particle C, we can use Coulomb's law:
Coulomb's law states that the magnitude of the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:
F = k * |q1 * q2| / r^2
where F is the magnitude of the force, k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between them.
Let's calculate the magnitude and direction of the resultant electric force on particle C:
1. Calculate the distances from particle C to A and B:
- From particle A to C: r_AC = √((x_C - x_A)^2 + (y_C - y_A)^2)
- From particle B to C: r_BC = √((x_C - x_B)^2 + (y_C - y_B)^2)
Given:
- Particle A charge, q_A = 3.00 x 10^-4 C
- Particle B charge, q_B = -6.00 x 10^-4 C
- Particle C charge, q_C = 1.00 x 10^-4 C
- Particle A position, (x_A, y_A) = (0, 0)
- Particle B position, (x_B, y_B) = (4.00 m, 0)
- Particle C position, (x_C, y_C) = (0, 3.00 m)
Plugging in the values:
- r_AC = √((0 - 0)^2 + (3.00 - 0)^2) = √(0 + 9.00) = √9.00 = 3.00 m
- r_BC = √((0 - 4.00)^2 + (3.00 - 0)^2) = √((-4.00)^2 + 9.00) = √16.00 + 9.00 = √25.00 = 5.00 m
2. Calculate the magnitude of the forces between each pair of charges:
- Force between A and C: F_AC = k * |q_A * q_C| / r_AC^2
- Force between B and C: F_BC = k * |q_B * q_C| / r_BC^2
Plugging in the values:
- F_AC = (9.0 x 10^9 N m^2/C^2) * |(3.00 x 10^-4 C) * (1.00 x 10^-4 C)| / (3.00)^2 = (9.0 x 10^9 N m^2/C^2) * (3.00 x 10^-8 C^2) / 9.00 = 3.0 x 10^-8 N
- F_BC = (9.0 x 10^9 N m^2/C^2) * |(-6.00 x 10^-4 C) * (1.00 x 10^-4 C)| / (5.00)^2 = (9.0 x 10^9 N m^2/C^2) * (6.00 x 10^-10 C^2) / 25.00 = 2.16 x 10^-10 N
3. Calculate the vector components of the forces:
- Force F_AC has a magnitude of 3.0 x 10^-8 N and acts along the line connecting A and C.
- Force F_BC has a magnitude of 2.16 x 10^-10 N and acts along the line connecting B and C.
4. Add the vector components to find the resultant force:
- Add the x-components of the forces: Fx_res = F_AC_cos(θ_AC) + F_BC_x
- Add the y-components of the forces: Fy_res = F_AC_sin(θ_AC) + F_BC_y
5. Calculate the magnitude and direction of the resultant force:
- Magnitude: F_res = √(Fx_res^2 + Fy_res^2)
- Direction: θ_res = atan2(Fy_res, Fx_res), which gives the angle relative to the positive x-axis.
By following these steps, you should be able to calculate the magnitude and direction of the resultant electric force acting on particle C.