\(\frac{m\ g}{q \ sin \theta}\)the equilibrium condition and the electric force we can find that the results are:
a) Electric field E= 9.33 10² N/C
b) String tension T = 3.39 109-2 N
Given parameters.
- Body mass m= 2.9 g = 2.9 10-3 kg.
- Electric charge of the body Q = 42 mC = 42 10-6 C.
To find.
a) Electric field.
b) String tension.
Equilibrium Condition.
Newton's second law gives a relationship between the net force, the mass and the acceleration of the body, in the case that the acceleration is zero it is called a condition of equilibrium.
∑ F = 0
A free body diagram is a diagram of the forces of the system without the details of the bodies, in the attachment we see a free body diagram of the system.
They indicate that the body is in an electric field, therefore the electric force is
F = q E
Where q is the charge of the body E the electric field that is a vectorial magnitude.
Let us write the equilibrium condition for each axis.
y-axis
Fe sin 30 -W = 0
q E sin 30 = m g
E = \(\frac{m \g }{q \ sin \theta }\)
Let's calculate.
E = \(\frac { 2.9 \ 10^{-3} \ 9.8 }{42 \ 10^{-6} \ sin30 }\)
E = 9.33 10² N/C
x-axis
Fe cos 30 – T = 0
T = qE cos 30
Let's calculate.
T = 42 10-6 9.33 10² cos 30
T = 3.39 10-2N
In conclusion using the equilibrium condition and the electric force we can find that the results are:
a) Electric field E= 9.33 10² N/C
b) String tension T = 3.39 10³ N
Learn more about the equilibrium condition here: brainly.com/question/24742518
