An object of mass m= 4.8g and charge Q = 49uC is attached to a string and placed in a uniform electric field that is inclined at an angle of 30.0 with the horizontal (see the figure). 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
To find the magnitude of the electric field (E) and the tension in the string, we can break down the problem into two parts:
1) Finding the magnitude of the electric field (E):
The object is in static equilibrium, which means that the net force acting on it is zero. In this case, the weight of the object (mg) and the tension in the string (T) will balance with the electric force (Eq) due to the electric field.
Since the object is in equilibrium, the vertical components of the forces must balance individually, and the horizontal components must balance as well.
For the vertical forces:
T * cos(θ) = mg -- Equation 1
For the horizontal forces:
T * sin(θ) = Eq -- Equation 2
Since the object is in static equilibrium, we know that mg = T as the object is attached to a string and is not accelerating vertically.
Substituting mg with T in Equation 1:
T * cos(θ) = T
Dividing both sides by T, we get:
cos(θ) = 1
Since cos(θ) = 1, we know that θ = 0°.
Now, we can use Equation 2 to find the magnitude of the electric field E:
T * sin(θ) = Eq
T * sin(0°) = Eq
0 = Eq
Since sin(0°) = 0, we know that Eq = 0.
Therefore, the magnitude of the electric field (E) is zero.
a) The magnitude of the electric field is zero.
2) Finding the tension in the string (T):
We can use Equation 1 to find the tension in the string (T):
T * cos(θ) = mg
Substituting the given values:
T * cos(30°) = (4.8g) * (9.8 m/s^2)
Converting the mass from grams to kilograms:
T * cos(30°) = (0.0048 kg) * (9.8 m/s^2)
Simplifying:
T * 0.866 = 0.04704 kg*m/s^2
Dividing both sides by 0.866:
T = 0.05435 kg*m/s^2
b) The tension in the string is approximately 0.05435 kg*m/s^2.