A small object of mass 3.00 g and charge -13 µC "floats" in a uniform electric field. What is the magnitude and direction of the electric field?
This may seem tricky but the key to realize is that electric forces and gravitational forces = 0. In this case Mg=Eq since the charge floats in the electric field. there the equation = E=mg/q and that'll be your answer. (You can solve the math).
To find the magnitude and direction of the electric field, we need to use the formula for the electric force on a charged particle in an electric field:
F = qE
where:
- F is the electric force,
- q is the charge of the object, and
- E is the electric field.
In this case, the object has a charge of -13 µC and it "floats" in the electric field, which means the electric force and the gravitational force balance each other. Therefore, we can set the magnitude of the electric force equal to the magnitude of the gravitational force:
|F| = mg
where:
- m is the mass of the object, and
- g is the acceleration due to gravity.
We can rearrange this equation to find the magnitude of the electric field:
E = |F|/q = mg/|q|
Plugging in the values given:
- m = 3.00 g = 0.00300 kg (converting grams to kilograms)
- g = 9.8 m/s^2 (acceleration due to gravity)
- q = -13 µC = -13 * 10^-6 C (converting microcoulombs to coulombs)
E = (0.00300 kg * 9.8 m/s^2) / (|-13 * 10^-6 C|)
E = (0.0294 kg m/s^2) / (13 * 10^-6 C)
E = 2.26 N/C
The magnitude of the electric field is 2.26 N/C. However, we also need to determine the direction of the electric field. Since the charge of the object is negative (-13 µC), the electric field will be directed in the opposite direction of the force experienced by a positive charge. Therefore, the electric field in this case is directed upward.