An object of mass m = 2.9 g and charge Q = +42 µC is attached to a string and placed in a uniform electric field that is inclined at an angle of 30.0° with the horizontal. The object is in static equilibrium when the string is horizontal.
a) Find the magnitude of the electric field.
b) Find the tension in the string.
i know A is -mg + sin 30 * qE =0 but i don't know how to solve for it
Eq sin 30 = mg (make sure to change mass to kg). Find E. Now Eq cos 30 = T. A Free Body Diagram is most helpful in this problem.
To solve for the magnitude of the electric field (E) and the tension in the string, you can use the given information and apply the principles of static equilibrium.
a) To find the magnitude of the electric field, we can start by analyzing the forces acting on the object attached to the string. In this case, the relevant forces are the gravitational force (mg) and the electrostatic force (qE), where q is the charge of the object.
In static equilibrium, the sum of the forces in both the x-direction and y-direction must be zero. Given that the string is horizontal, we only need to consider forces in the vertical (y-) direction.
The force equation for the y-direction can be written as:
-F_gravity + F_electric = 0
Substituting the expressions for these forces, we have:
-mg + qE = 0
Rearranging the equation, we can solve for the electric field (E):
E = mg / q
Plugging in the given values:
m = 2.9 g = 0.0029 kg
q = +42 µC = 42 x 10^(-6) C
g = acceleration due to gravity = 9.8 m/s^2
E = (0.0029 kg) x (9.8 m/s^2) / (42 x 10^(-6) C)
E ≈ 0.676 N/C
Therefore, the magnitude of the electric field is approximately 0.676 N/C.
b) To find the tension in the string, we need to consider the forces acting in the horizontal (x-) direction. Since the object is in static equilibrium, the tension in the string must balance the horizontal component of the electrostatic force.
The force equation for the x-direction can be written as:
F_tension = F_electric * cosθ
Where θ is the angle between the electric field and the horizontal (30° in this case).
Substituting the expressions for these forces, we have:
T = qE * cosθ
Plugging in the given values:
q = +42 µC = 42 x 10^(-6) C
E ≈ 0.676 N/C
θ = 30°
T = (42 x 10^(-6) C) * (0.676 N/C) * cos(30°)
T ≈ 0.022 N
Therefore, the tension in the string is approximately 0.022 N.